Why is 2 * (i * i) faster than 2 * i * i in Java?
The following Java program takes on average between 0.50s and 0.55s to run:
public static void main(String args) {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
System.out.println((double) (System.nanoTime() - startTime) / 1000000000 + " s");
System.out.println("n = " + n);
}
If I replace 2 * (i * i)
with 2 * i * i
, it takes between 0.60 and 0.65s to run. How come?
I ran each version of the program 15 times, alternating between the two. Here are the results:
2*(i*i) | 2*i*i
----------+----------
0.5183738 | 0.6246434
0.5298337 | 0.6049722
0.5308647 | 0.6603363
0.5133458 | 0.6243328
0.5003011 | 0.6541802
0.5366181 | 0.6312638
0.515149 | 0.6241105
0.5237389 | 0.627815
0.5249942 | 0.6114252
0.5641624 | 0.6781033
0.538412 | 0.6393969
0.5466744 | 0.6608845
0.531159 | 0.6201077
0.5048032 | 0.6511559
0.5232789 | 0.6544526
The fastest run of 2 * i * i
took longer than the slowest run of 2 * (i * i)
. If they were both as efficient, the probability of this happening would be less than 1/2^15 = 0.00305%.
java performance benchmarking bytecode jit
|
show 12 more comments
The following Java program takes on average between 0.50s and 0.55s to run:
public static void main(String args) {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
System.out.println((double) (System.nanoTime() - startTime) / 1000000000 + " s");
System.out.println("n = " + n);
}
If I replace 2 * (i * i)
with 2 * i * i
, it takes between 0.60 and 0.65s to run. How come?
I ran each version of the program 15 times, alternating between the two. Here are the results:
2*(i*i) | 2*i*i
----------+----------
0.5183738 | 0.6246434
0.5298337 | 0.6049722
0.5308647 | 0.6603363
0.5133458 | 0.6243328
0.5003011 | 0.6541802
0.5366181 | 0.6312638
0.515149 | 0.6241105
0.5237389 | 0.627815
0.5249942 | 0.6114252
0.5641624 | 0.6781033
0.538412 | 0.6393969
0.5466744 | 0.6608845
0.531159 | 0.6201077
0.5048032 | 0.6511559
0.5232789 | 0.6544526
The fastest run of 2 * i * i
took longer than the slowest run of 2 * (i * i)
. If they were both as efficient, the probability of this happening would be less than 1/2^15 = 0.00305%.
java performance benchmarking bytecode jit
4
I get similar results (slightly different numbers, but definitely noticeable and consistent gap, definitely more than sampling error)
– Krease
Nov 23 at 20:47
27
Also please see: stackoverflow.com/questions/504103/…
– lexicore
Nov 23 at 20:56
2
@Krease Good that you caught my mistake. According to the new benchmark I ran2 * i * i
is slower. I'll try running with Graal as well.
– Jorn Vernee
Nov 23 at 21:07
5
@nullpointer To find out for real why one is faster than the other, we'd have to get the disassembly or Ideal graphs for those methods. The assembler is very annoying to try and figure out, so I'm trying to get an OpenJDK debug build which can output nice graphs.
– Jorn Vernee
Nov 23 at 21:29
4
You could rename your question to "Why isi * i * 2
faster than2 * i * i
?" for improved clarity that the issue is on the order of the operations.
– Cœur
Nov 28 at 4:02
|
show 12 more comments
The following Java program takes on average between 0.50s and 0.55s to run:
public static void main(String args) {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
System.out.println((double) (System.nanoTime() - startTime) / 1000000000 + " s");
System.out.println("n = " + n);
}
If I replace 2 * (i * i)
with 2 * i * i
, it takes between 0.60 and 0.65s to run. How come?
I ran each version of the program 15 times, alternating between the two. Here are the results:
2*(i*i) | 2*i*i
----------+----------
0.5183738 | 0.6246434
0.5298337 | 0.6049722
0.5308647 | 0.6603363
0.5133458 | 0.6243328
0.5003011 | 0.6541802
0.5366181 | 0.6312638
0.515149 | 0.6241105
0.5237389 | 0.627815
0.5249942 | 0.6114252
0.5641624 | 0.6781033
0.538412 | 0.6393969
0.5466744 | 0.6608845
0.531159 | 0.6201077
0.5048032 | 0.6511559
0.5232789 | 0.6544526
The fastest run of 2 * i * i
took longer than the slowest run of 2 * (i * i)
. If they were both as efficient, the probability of this happening would be less than 1/2^15 = 0.00305%.
java performance benchmarking bytecode jit
The following Java program takes on average between 0.50s and 0.55s to run:
public static void main(String args) {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
System.out.println((double) (System.nanoTime() - startTime) / 1000000000 + " s");
System.out.println("n = " + n);
}
If I replace 2 * (i * i)
with 2 * i * i
, it takes between 0.60 and 0.65s to run. How come?
I ran each version of the program 15 times, alternating between the two. Here are the results:
2*(i*i) | 2*i*i
----------+----------
0.5183738 | 0.6246434
0.5298337 | 0.6049722
0.5308647 | 0.6603363
0.5133458 | 0.6243328
0.5003011 | 0.6541802
0.5366181 | 0.6312638
0.515149 | 0.6241105
0.5237389 | 0.627815
0.5249942 | 0.6114252
0.5641624 | 0.6781033
0.538412 | 0.6393969
0.5466744 | 0.6608845
0.531159 | 0.6201077
0.5048032 | 0.6511559
0.5232789 | 0.6544526
The fastest run of 2 * i * i
took longer than the slowest run of 2 * (i * i)
. If they were both as efficient, the probability of this happening would be less than 1/2^15 = 0.00305%.
java performance benchmarking bytecode jit
java performance benchmarking bytecode jit
edited Nov 30 at 20:21
Alex Riley
77.1k21156161
77.1k21156161
asked Nov 23 at 20:40
Stefan
2,692339
2,692339
4
I get similar results (slightly different numbers, but definitely noticeable and consistent gap, definitely more than sampling error)
– Krease
Nov 23 at 20:47
27
Also please see: stackoverflow.com/questions/504103/…
– lexicore
Nov 23 at 20:56
2
@Krease Good that you caught my mistake. According to the new benchmark I ran2 * i * i
is slower. I'll try running with Graal as well.
– Jorn Vernee
Nov 23 at 21:07
5
@nullpointer To find out for real why one is faster than the other, we'd have to get the disassembly or Ideal graphs for those methods. The assembler is very annoying to try and figure out, so I'm trying to get an OpenJDK debug build which can output nice graphs.
– Jorn Vernee
Nov 23 at 21:29
4
You could rename your question to "Why isi * i * 2
faster than2 * i * i
?" for improved clarity that the issue is on the order of the operations.
– Cœur
Nov 28 at 4:02
|
show 12 more comments
4
I get similar results (slightly different numbers, but definitely noticeable and consistent gap, definitely more than sampling error)
– Krease
Nov 23 at 20:47
27
Also please see: stackoverflow.com/questions/504103/…
– lexicore
Nov 23 at 20:56
2
@Krease Good that you caught my mistake. According to the new benchmark I ran2 * i * i
is slower. I'll try running with Graal as well.
– Jorn Vernee
Nov 23 at 21:07
5
@nullpointer To find out for real why one is faster than the other, we'd have to get the disassembly or Ideal graphs for those methods. The assembler is very annoying to try and figure out, so I'm trying to get an OpenJDK debug build which can output nice graphs.
– Jorn Vernee
Nov 23 at 21:29
4
You could rename your question to "Why isi * i * 2
faster than2 * i * i
?" for improved clarity that the issue is on the order of the operations.
– Cœur
Nov 28 at 4:02
4
4
I get similar results (slightly different numbers, but definitely noticeable and consistent gap, definitely more than sampling error)
– Krease
Nov 23 at 20:47
I get similar results (slightly different numbers, but definitely noticeable and consistent gap, definitely more than sampling error)
– Krease
Nov 23 at 20:47
27
27
Also please see: stackoverflow.com/questions/504103/…
– lexicore
Nov 23 at 20:56
Also please see: stackoverflow.com/questions/504103/…
– lexicore
Nov 23 at 20:56
2
2
@Krease Good that you caught my mistake. According to the new benchmark I ran
2 * i * i
is slower. I'll try running with Graal as well.– Jorn Vernee
Nov 23 at 21:07
@Krease Good that you caught my mistake. According to the new benchmark I ran
2 * i * i
is slower. I'll try running with Graal as well.– Jorn Vernee
Nov 23 at 21:07
5
5
@nullpointer To find out for real why one is faster than the other, we'd have to get the disassembly or Ideal graphs for those methods. The assembler is very annoying to try and figure out, so I'm trying to get an OpenJDK debug build which can output nice graphs.
– Jorn Vernee
Nov 23 at 21:29
@nullpointer To find out for real why one is faster than the other, we'd have to get the disassembly or Ideal graphs for those methods. The assembler is very annoying to try and figure out, so I'm trying to get an OpenJDK debug build which can output nice graphs.
– Jorn Vernee
Nov 23 at 21:29
4
4
You could rename your question to "Why is
i * i * 2
faster than 2 * i * i
?" for improved clarity that the issue is on the order of the operations.– Cœur
Nov 28 at 4:02
You could rename your question to "Why is
i * i * 2
faster than 2 * i * i
?" for improved clarity that the issue is on the order of the operations.– Cœur
Nov 28 at 4:02
|
show 12 more comments
10 Answers
10
active
oldest
votes
There is a slight difference in the ordering of the bytecode.
2 * (i * i)
:
iconst_2
iload0
iload0
imul
imul
iadd
vs 2 * i * i
:
iconst_2
iload0
imul
iload0
imul
iadd
At first sight this should not make a difference; if anything the second version is more optimal since it uses one slot less.
So we need to dig deeper into the lower level (JIT)1.
Remember that JIT tends to unroll small loops very aggressively. Indeed we observe a 16x unrolling for the 2 * (i * i)
case:
030 B2: # B2 B3 <- B1 B2 Loop: B2-B2 inner main of N18 Freq: 1e+006
030 addl R11, RBP # int
033 movl RBP, R13 # spill
036 addl RBP, #14 # int
039 imull RBP, RBP # int
03c movl R9, R13 # spill
03f addl R9, #13 # int
043 imull R9, R9 # int
047 sall RBP, #1
049 sall R9, #1
04c movl R8, R13 # spill
04f addl R8, #15 # int
053 movl R10, R8 # spill
056 movdl XMM1, R8 # spill
05b imull R10, R8 # int
05f movl R8, R13 # spill
062 addl R8, #12 # int
066 imull R8, R8 # int
06a sall R10, #1
06d movl [rsp + #32], R10 # spill
072 sall R8, #1
075 movl RBX, R13 # spill
078 addl RBX, #11 # int
07b imull RBX, RBX # int
07e movl RCX, R13 # spill
081 addl RCX, #10 # int
084 imull RCX, RCX # int
087 sall RBX, #1
089 sall RCX, #1
08b movl RDX, R13 # spill
08e addl RDX, #8 # int
091 imull RDX, RDX # int
094 movl RDI, R13 # spill
097 addl RDI, #7 # int
09a imull RDI, RDI # int
09d sall RDX, #1
09f sall RDI, #1
0a1 movl RAX, R13 # spill
0a4 addl RAX, #6 # int
0a7 imull RAX, RAX # int
0aa movl RSI, R13 # spill
0ad addl RSI, #4 # int
0b0 imull RSI, RSI # int
0b3 sall RAX, #1
0b5 sall RSI, #1
0b7 movl R10, R13 # spill
0ba addl R10, #2 # int
0be imull R10, R10 # int
0c2 movl R14, R13 # spill
0c5 incl R14 # int
0c8 imull R14, R14 # int
0cc sall R10, #1
0cf sall R14, #1
0d2 addl R14, R11 # int
0d5 addl R14, R10 # int
0d8 movl R10, R13 # spill
0db addl R10, #3 # int
0df imull R10, R10 # int
0e3 movl R11, R13 # spill
0e6 addl R11, #5 # int
0ea imull R11, R11 # int
0ee sall R10, #1
0f1 addl R10, R14 # int
0f4 addl R10, RSI # int
0f7 sall R11, #1
0fa addl R11, R10 # int
0fd addl R11, RAX # int
100 addl R11, RDI # int
103 addl R11, RDX # int
106 movl R10, R13 # spill
109 addl R10, #9 # int
10d imull R10, R10 # int
111 sall R10, #1
114 addl R10, R11 # int
117 addl R10, RCX # int
11a addl R10, RBX # int
11d addl R10, R8 # int
120 addl R9, R10 # int
123 addl RBP, R9 # int
126 addl RBP, [RSP + #32 (32-bit)] # int
12a addl R13, #16 # int
12e movl R11, R13 # spill
131 imull R11, R13 # int
135 sall R11, #1
138 cmpl R13, #999999985
13f jl B2 # loop end P=1.000000 C=6554623.000000
We see that there is 1 register that is "spilled" onto the stack.
And for the 2 * i * i
version:
05a B3: # B2 B4 <- B1 B2 Loop: B3-B2 inner main of N18 Freq: 1e+006
05a addl RBX, R11 # int
05d movl [rsp + #32], RBX # spill
061 movl R11, R8 # spill
064 addl R11, #15 # int
068 movl [rsp + #36], R11 # spill
06d movl R11, R8 # spill
070 addl R11, #14 # int
074 movl R10, R9 # spill
077 addl R10, #16 # int
07b movdl XMM2, R10 # spill
080 movl RCX, R9 # spill
083 addl RCX, #14 # int
086 movdl XMM1, RCX # spill
08a movl R10, R9 # spill
08d addl R10, #12 # int
091 movdl XMM4, R10 # spill
096 movl RCX, R9 # spill
099 addl RCX, #10 # int
09c movdl XMM6, RCX # spill
0a0 movl RBX, R9 # spill
0a3 addl RBX, #8 # int
0a6 movl RCX, R9 # spill
0a9 addl RCX, #6 # int
0ac movl RDX, R9 # spill
0af addl RDX, #4 # int
0b2 addl R9, #2 # int
0b6 movl R10, R14 # spill
0b9 addl R10, #22 # int
0bd movdl XMM3, R10 # spill
0c2 movl RDI, R14 # spill
0c5 addl RDI, #20 # int
0c8 movl RAX, R14 # spill
0cb addl RAX, #32 # int
0ce movl RSI, R14 # spill
0d1 addl RSI, #18 # int
0d4 movl R13, R14 # spill
0d7 addl R13, #24 # int
0db movl R10, R14 # spill
0de addl R10, #26 # int
0e2 movl [rsp + #40], R10 # spill
0e7 movl RBP, R14 # spill
0ea addl RBP, #28 # int
0ed imull RBP, R11 # int
0f1 addl R14, #30 # int
0f5 imull R14, [RSP + #36 (32-bit)] # int
0fb movl R10, R8 # spill
0fe addl R10, #11 # int
102 movdl R11, XMM3 # spill
107 imull R11, R10 # int
10b movl [rsp + #44], R11 # spill
110 movl R10, R8 # spill
113 addl R10, #10 # int
117 imull RDI, R10 # int
11b movl R11, R8 # spill
11e addl R11, #8 # int
122 movdl R10, XMM2 # spill
127 imull R10, R11 # int
12b movl [rsp + #48], R10 # spill
130 movl R10, R8 # spill
133 addl R10, #7 # int
137 movdl R11, XMM1 # spill
13c imull R11, R10 # int
140 movl [rsp + #52], R11 # spill
145 movl R11, R8 # spill
148 addl R11, #6 # int
14c movdl R10, XMM4 # spill
151 imull R10, R11 # int
155 movl [rsp + #56], R10 # spill
15a movl R10, R8 # spill
15d addl R10, #5 # int
161 movdl R11, XMM6 # spill
166 imull R11, R10 # int
16a movl [rsp + #60], R11 # spill
16f movl R11, R8 # spill
172 addl R11, #4 # int
176 imull RBX, R11 # int
17a movl R11, R8 # spill
17d addl R11, #3 # int
181 imull RCX, R11 # int
185 movl R10, R8 # spill
188 addl R10, #2 # int
18c imull RDX, R10 # int
190 movl R11, R8 # spill
193 incl R11 # int
196 imull R9, R11 # int
19a addl R9, [RSP + #32 (32-bit)] # int
19f addl R9, RDX # int
1a2 addl R9, RCX # int
1a5 addl R9, RBX # int
1a8 addl R9, [RSP + #60 (32-bit)] # int
1ad addl R9, [RSP + #56 (32-bit)] # int
1b2 addl R9, [RSP + #52 (32-bit)] # int
1b7 addl R9, [RSP + #48 (32-bit)] # int
1bc movl R10, R8 # spill
1bf addl R10, #9 # int
1c3 imull R10, RSI # int
1c7 addl R10, R9 # int
1ca addl R10, RDI # int
1cd addl R10, [RSP + #44 (32-bit)] # int
1d2 movl R11, R8 # spill
1d5 addl R11, #12 # int
1d9 imull R13, R11 # int
1dd addl R13, R10 # int
1e0 movl R10, R8 # spill
1e3 addl R10, #13 # int
1e7 imull R10, [RSP + #40 (32-bit)] # int
1ed addl R10, R13 # int
1f0 addl RBP, R10 # int
1f3 addl R14, RBP # int
1f6 movl R10, R8 # spill
1f9 addl R10, #16 # int
1fd cmpl R10, #999999985
204 jl B2 # loop end P=1.000000 C=7419903.000000
Here we observe much more "spilling" and more accesses to the stack [RSP + ...]
, due to more intermediate results that need to be preserved.
Thus the answer to the question is simple: 2 * (i * i)
is faster than 2 * i * i
because the JIT generates more optimal assembly code for the first case.
But of course it is obvious that neither the first nor the second version is any good; the loop could really benefit from vectorization, since any x86-64 CPU has at least SSE2 support.
So it's an issue of the optimizer; as is often the case, it unrolls too aggressively and shoots itself in the foot, all the while missing out on various other opportunities.
In fact, modern x86-64 CPUs break down the instructions further into micro-ops (µops) and with features like register renaming, µop caches and loop buffers, loop optimization takes a lot more finesse than a simple unrolling for optimal performance. According to Agner Fog's optimization guide:
The gain in performance due to the µop cache can be quite
considerable if the average instruction length is more than 4 bytes.
The following methods of optimizing the use of the µop cache may
be considered:
- Make sure that critical loops are small enough to fit into the µop cache.
- Align the most critical loop entries and function entries by 32.
- Avoid unnecessary loop unrolling.
- Avoid instructions that have extra load time
. . .
Regarding those load times - even the fastest L1D hit costs 4 cycles, an extra register and µop, so yes, even a few accesses to memory will hurt performance in tight loops.
But back to the vectorization opportunity - to see how fast it can be, we can compile a similar C application with GCC, which outright vectorizes it (AVX2 is shown, SSE2 is similar)2:
vmovdqa ymm0, YMMWORD PTR .LC0[rip]
vmovdqa ymm3, YMMWORD PTR .LC1[rip]
xor eax, eax
vpxor xmm2, xmm2, xmm2
.L2:
vpmulld ymm1, ymm0, ymm0
inc eax
vpaddd ymm0, ymm0, ymm3
vpslld ymm1, ymm1, 1
vpaddd ymm2, ymm2, ymm1
cmp eax, 125000000 ; 8 calculations per iteration
jne .L2
vmovdqa xmm0, xmm2
vextracti128 xmm2, ymm2, 1
vpaddd xmm2, xmm0, xmm2
vpsrldq xmm0, xmm2, 8
vpaddd xmm0, xmm2, xmm0
vpsrldq xmm1, xmm0, 4
vpaddd xmm0, xmm0, xmm1
vmovd eax, xmm0
vzeroupper
With run times:
- SSE: 0.24 s, or 2 times faster.
- AVX: 0.15 s, or 3 times faster.
- AVX2: 0.08 s, or 5 times faster.
1To get JIT generated assembly output, get a debug JVM and run with -XX:+PrintOptoAssembly
2The C version is compiled with the -fwrapv
flag, which enables GCC to treat signed integer overflow as a two's-complement wrap-around.
10
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
36
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
12
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just aret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.
– Peter Cordes
Nov 25 at 22:26
8
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole formov
/add-immediate
. e.g.movl RBX, R9
/addl RBX, #8
should beleal ebx, [r9 + 8]
, 1 uop to copy-and-add. Orleal ebx, [r9 + r9 + 16]
to doebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.
– Peter Cordes
Nov 25 at 22:38
6
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
|
show 18 more comments
When the multiplication is 2 * (i * i)
, the JVM is able to factor out the multiplication by 2
from the loop, resulting in this equivalent but more efficient code:
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
but when the multiplication is (2 * i) * i
, the JVM doesn't optimize it since the multiplication by a constant is no longer right before the addition.
Here are a few reasons why I think this is the case:
- Adding an
if (n == 0) n = 1
statement at the start of the loop results in both versions being as efficient, since factoring out the multiplication no longer guarantees that the result will be the same - The optimized version (by factoring out the multiplication by 2) is exactly as fast as the
2 * (i * i)
version
Here is the test code that I used to draw these conclusions:
public static void main(String args) {
long fastVersion = 0;
long slowVersion = 0;
long optimizedVersion = 0;
long modifiedFastVersion = 0;
long modifiedSlowVersion = 0;
for (int i = 0; i < 10; i++) {
fastVersion += fastVersion();
slowVersion += slowVersion();
optimizedVersion += optimizedVersion();
modifiedFastVersion += modifiedFastVersion();
modifiedSlowVersion += modifiedSlowVersion();
}
System.out.println("Fast version: " + (double) fastVersion / 1000000000 + " s");
System.out.println("Slow version: " + (double) slowVersion / 1000000000 + " s");
System.out.println("Optimized version: " + (double) optimizedVersion / 1000000000 + " s");
System.out.println("Modified fast version: " + (double) modifiedFastVersion / 1000000000 + " s");
System.out.println("Modified slow version: " + (double) modifiedSlowVersion / 1000000000 + " s");
}
private static long fastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long slowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
private static long optimizedVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
return System.nanoTime() - startTime;
}
private static long modifiedFastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long modifiedSlowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
And here are the results:
Fast version: 5.7274411 s
Slow version: 7.6190804 s
Optimized version: 5.1348007 s
Modified fast version: 7.1492705 s
Modified slow version: 7.2952668 s
3
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
2
I think on the optimizedVersion, it should ben *= 2000000000;
– StefansArya
Nov 24 at 1:19
4
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.
– Martin Bonner
Nov 26 at 15:57
5
The math equation was just like this2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.
– StefansArya
Nov 26 at 17:22
2
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
|
show 2 more comments
ByteCodes: https://cs.nyu.edu/courses/fall00/V22.0201-001/jvm2.html
ByteCodes Viewer: https://github.com/Konloch/bytecode-viewer
On my JDK (Win10 64 1.8.0_65-b17) I can reproduce and explain:
public static void main(String args) {
int repeat = 10;
long A = 0;
long B = 0;
for (int i = 0; i < repeat; i++) {
A += test();
B += testB();
}
System.out.println(A / repeat + " ms");
System.out.println(B / repeat + " ms");
}
private static long test() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multi(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multi(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms A " + n);
return ms;
}
private static long testB() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multiB(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multiB(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms B " + n);
return ms;
}
private static int multiB(int i) {
return 2 * (i * i);
}
private static int multi(int i) {
return 2 * i * i;
}
Output:
...
405 ms A 785527736
327 ms B 785527736
404 ms A 785527736
329 ms B 785527736
404 ms A 785527736
328 ms B 785527736
404 ms A 785527736
328 ms B 785527736
410 ms
333 ms
So why?
The Byte code is this:
private static multiB(int arg0) { // 2 * (i * i)
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
iload0
imul
imul
ireturn
}
L2 {
}
}
private static multi(int arg0) { // 2 * i * i
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
imul
iload0
imul
ireturn
}
L2 {
}
}
The difference being:
With brackets (2 * (i * i)
):
- push const stack
- push local on stack
- push local on stack
- multiply top of stack
- multiply top of stack
Without brackets (2 * i * i
):
- push const stack
- push local on stack
- multiply top of stack
- push local on stack
- multiply top of stack
Loading all on stack and then working back down is faster than switching between putting on stack and operating on it.
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
add a comment |
Kasperd asked in a comment of the accepted answer:
The Java and C examples use quite different register names. Are both example using the AMD64 ISA?
xor edx, edx
xor eax, eax
.L2:
mov ecx, edx
imul ecx, edx
add edx, 1
lea eax, [rax+rcx*2]
cmp edx, 1000000000
jne .L2
I don't have enough reputation to answer this in the comments, but these are the same ISA. It's worth pointing out that the GCC version uses 32-bit integer logic and the JVM compiled version uses 64-bit integer logic internally.
R8 to R15 are just new X86_64 registers. EAX to EDX are the lower parts of the RAX to RDX general purpose registers. The important part in the answer is that the GCC version is not unrolled. It simply executes one round of the loop per actual machine code loop. While the JVM version has 16 rounds of the loop in one physical loop (based on rustyx answer, I did not reinterpret the assembly). This is one of the reasons why there are more registers being used since the loop body is actually 16 times longer.
add a comment |
While not directly related to the question's environment, just for the curiosity, I did the same test on .Net Core 2.1, x64, release mode.
Here is the interesting result, confirming similar phonemenia (other way around) happening over the dark side of the force. Code:
static void Main(string args)
{
Stopwatch watch = new Stopwatch();
Console.WriteLine("2 * (i * i)");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * (i * i);
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
Console.WriteLine();
Console.WriteLine("2 * i * i");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * i * i;
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
}
Result:
2 * (i * i)
- result:119860736, 438ms
- result:119860736, 433ms
- result:119860736, 437ms
- result:119860736, 435ms
- result:119860736, 436ms
- result:119860736, 435ms
- result:119860736, 435ms
- result:119860736, 439ms
- result:119860736, 436ms
- result:119860736, 437ms
2 * i * i
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 418ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 416ms
- result:119860736, 417ms
- result:119860736, 418ms
1
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
1
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
1
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
5
Except this is the other way around
– leppie
Nov 30 at 14:55
2
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
|
show 4 more comments
I got similar results:
2 * (i * i): 0.458765943 s, n=119860736
2 * i * i: 0.580255126 s, n=119860736
I got the SAME results if both loops were in the same program, or each was in a separate .java file/.class, executed on a separate run.
Finally, here is a javap -c -v <.java>
decompile of each:
3: ldc #3 // String 2 * (i * i):
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: iload 4
30: imul
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
vs.
3: ldc #3 // String 2 * i * i:
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: imul
29: iload 4
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
FYI -
java -version
java version "1.8.0_121"
Java(TM) SE Runtime Environment (build 1.8.0_121-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.121-b13, mixed mode)
1
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
2
Get a debug jre and run with-XX:+PrintOptoAssembly
. Or just use vtune or alike.
– rustyx
Nov 23 at 22:42
1
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
|
show 2 more comments
I tried a JMH using the default archetype: I also added optimized version based Runemoro' explanation .
@State(Scope.Benchmark)
@Warmup(iterations = 2)
@Fork(1)
@Measurement(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
//@BenchmarkMode({ Mode.All })
@BenchmarkMode(Mode.AverageTime)
public class MyBenchmark {
@Param({ "100", "1000", "1000000000" })
private int size;
@Benchmark
public int two_square_i() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
@Benchmark
public int square_i_two() {
int n = 0;
for (int i = 0; i < size; i++) {
n += i * i;
}
return 2*n;
}
@Benchmark
public int two_i_() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
}
The result are here:
Benchmark (size) Mode Samples Score Score error Units
o.s.MyBenchmark.square_i_two 100 avgt 10 58,062 1,410 ns/op
o.s.MyBenchmark.square_i_two 1000 avgt 10 547,393 12,851 ns/op
o.s.MyBenchmark.square_i_two 1000000000 avgt 10 540343681,267 16795210,324 ns/op
o.s.MyBenchmark.two_i_ 100 avgt 10 87,491 2,004 ns/op
o.s.MyBenchmark.two_i_ 1000 avgt 10 1015,388 30,313 ns/op
o.s.MyBenchmark.two_i_ 1000000000 avgt 10 967100076,600 24929570,556 ns/op
o.s.MyBenchmark.two_square_i 100 avgt 10 70,715 2,107 ns/op
o.s.MyBenchmark.two_square_i 1000 avgt 10 686,977 24,613 ns/op
o.s.MyBenchmark.two_square_i 1000000000 avgt 10 652736811,450 27015580,488 ns/op
On my PC (Core i7 860, doing nothing much apart reading on my smartphone):
n += i*i
thenn*2
is first
2 * (i * i)
is second.
The JVM is clearly not optimizing the same way than a human does (based on Runemoro answer).
Now then, reading bytecode: javap -c -v ./target/classes/org/sample/MyBenchmark.class
- Differences between 2*(i*i) (left) and 2*i*i (right) here: https://www.diffchecker.com/cvSFppWI
- Differences between 2*(i*i) and the optimized version here: https://www.diffchecker.com/I1XFu5dP
I am not expert on bytecode but we iload_2
before we imul
: that's probably where you get the difference: I can suppose that the JVM optimize reading i
twice (i
is already here, there is no need to load it again) whilst in the 2*i*i
it can't.
3
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
add a comment |
More of an addendum. I did repro the experiment using the latest Java 8 JVM from IBM:
java version "1.8.0_191"
Java(TM) 2 Runtime Environment, Standard Edition (IBM build 1.8.0_191-b12 26_Oct_2018_18_45 Mac OS X x64(SR5 FP25))
Java HotSpot(TM) 64-Bit Server VM (build 25.191-b12, mixed mode)
and this shows very similar results:
0.374653912 s
n = 119860736
0.447778698 s
n = 119860736
( second results using 2 * i * i ).
Interestingly enough, when running on the same machine, but using Oracle java:
Java version "1.8.0_181"
Java(TM) SE Runtime Environment (build 1.8.0_181-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.181-b13, mixed mode)
results are on average a bit slower:
0.414331815 s
n = 119860736
0.491430656 s
n = 119860736
Long story short: even the minor version number of HotSpot matter here, as subtle differences within the JIT implementation can have notable effects.
add a comment |
Interesting observation using Java 11 and switching off loop unrolling with the following VM option:
-XX:LoopUnrollLimit=0
The loop with the 2 * (i * i)
expression results in a more compact native code1:
L0001: add eax,r11d
inc r8d
mov r11d,r8d
imul r11d,r8d
shl r11d,1h
cmp r8d,r10d
jl L0001
in comparison with the 2 * i * i
version:
L0001: add eax,r11d
mov r11d,r8d
shl r11d,1h
add r11d,2h
inc r8d
imul r11d,r8d
cmp r8d,r10d
jl L0001
Java version:
java version "11" 2018-09-25
Java(TM) SE Runtime Environment 18.9 (build 11+28)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11+28, mixed mode)
Benchmark results:
Benchmark (size) Mode Cnt Score Error Units
LoopTest.fast 1000000000 avgt 5 694,868 ± 36,470 ms/op
LoopTest.slow 1000000000 avgt 5 769,840 ± 135,006 ms/op
Benchmark source code:
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@Warmup(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@State(Scope.Thread)
@Fork(1)
public class LoopTest {
@Param("1000000000") private int size;
public static void main(String args) throws RunnerException {
Options opt =
new OptionsBuilder().include(LoopTest.class.getSimpleName())
.jvmArgs("-XX:LoopUnrollLimit=0")
.build();
new Runner(opt).run();
}
@Benchmark
public int slow() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int fast() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
}
1 - VM options used: -XX:+UnlockDiagnosticVMOptions -XX:+PrintAssembly -XX:LoopUnrollLimit=0
2
Wow, that's some braindead asm. Instead of incrementingi
before copying it to calculate2*i
, it does it after so it needs an extraadd r11d,2
instruction. (Plus it misses theadd same,same
peephole instead ofshl
by 1 (add runs on more ports). It also misses an LEA peephole forx*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.
– Peter Cordes
Dec 2 at 2:50
2
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)
– Peter Cordes
Dec 2 at 2:51
add a comment |
The two methods of adding do generate slightly different byte code:
17: iconst_2
18: iload 4
20: iload 4
22: imul
23: imul
24: iadd
For 2 * (i * i)
vs:
17: iconst_2
18: iload 4
20: imul
21: iload 4
23: imul
24: iadd
For 2 * i * i
.
And when using a JMH benchmark like this:
@Warmup(iterations = 5, batchSize = 1)
@Measurement(iterations = 5, batchSize = 1)
@Fork(1)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@State(Scope.Benchmark)
public class MyBenchmark {
@Benchmark
public int noBrackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int brackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return n;
}
}
The difference is clear:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: <none>
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 380.889 ± 58.011 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 512.464 ± 11.098 ms/op
What you observe is correct, and not just an anomaly of your benchmarking style (i.e. no warmup, see How do I write a correct micro-benchmark in Java?)
Running again with Graal:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: -XX:+UnlockExperimentalVMOptions -XX:+EnableJVMCI -XX:+UseJVMCICompiler
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 335.100 ± 23.085 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 331.163 ± 50.670 ms/op
You see that the results are much closer, which makes sense, since Graal is an overall better performing, more modern, compiler.
So this is really just up to how well the JIT compiler is able to optimize a particular piece of code, and doesn't necessarily have a logical reason to it.
uhrrr.. yournoBrackets
is using brackets..
– Krease
Nov 23 at 20:55
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
add a comment |
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There is a slight difference in the ordering of the bytecode.
2 * (i * i)
:
iconst_2
iload0
iload0
imul
imul
iadd
vs 2 * i * i
:
iconst_2
iload0
imul
iload0
imul
iadd
At first sight this should not make a difference; if anything the second version is more optimal since it uses one slot less.
So we need to dig deeper into the lower level (JIT)1.
Remember that JIT tends to unroll small loops very aggressively. Indeed we observe a 16x unrolling for the 2 * (i * i)
case:
030 B2: # B2 B3 <- B1 B2 Loop: B2-B2 inner main of N18 Freq: 1e+006
030 addl R11, RBP # int
033 movl RBP, R13 # spill
036 addl RBP, #14 # int
039 imull RBP, RBP # int
03c movl R9, R13 # spill
03f addl R9, #13 # int
043 imull R9, R9 # int
047 sall RBP, #1
049 sall R9, #1
04c movl R8, R13 # spill
04f addl R8, #15 # int
053 movl R10, R8 # spill
056 movdl XMM1, R8 # spill
05b imull R10, R8 # int
05f movl R8, R13 # spill
062 addl R8, #12 # int
066 imull R8, R8 # int
06a sall R10, #1
06d movl [rsp + #32], R10 # spill
072 sall R8, #1
075 movl RBX, R13 # spill
078 addl RBX, #11 # int
07b imull RBX, RBX # int
07e movl RCX, R13 # spill
081 addl RCX, #10 # int
084 imull RCX, RCX # int
087 sall RBX, #1
089 sall RCX, #1
08b movl RDX, R13 # spill
08e addl RDX, #8 # int
091 imull RDX, RDX # int
094 movl RDI, R13 # spill
097 addl RDI, #7 # int
09a imull RDI, RDI # int
09d sall RDX, #1
09f sall RDI, #1
0a1 movl RAX, R13 # spill
0a4 addl RAX, #6 # int
0a7 imull RAX, RAX # int
0aa movl RSI, R13 # spill
0ad addl RSI, #4 # int
0b0 imull RSI, RSI # int
0b3 sall RAX, #1
0b5 sall RSI, #1
0b7 movl R10, R13 # spill
0ba addl R10, #2 # int
0be imull R10, R10 # int
0c2 movl R14, R13 # spill
0c5 incl R14 # int
0c8 imull R14, R14 # int
0cc sall R10, #1
0cf sall R14, #1
0d2 addl R14, R11 # int
0d5 addl R14, R10 # int
0d8 movl R10, R13 # spill
0db addl R10, #3 # int
0df imull R10, R10 # int
0e3 movl R11, R13 # spill
0e6 addl R11, #5 # int
0ea imull R11, R11 # int
0ee sall R10, #1
0f1 addl R10, R14 # int
0f4 addl R10, RSI # int
0f7 sall R11, #1
0fa addl R11, R10 # int
0fd addl R11, RAX # int
100 addl R11, RDI # int
103 addl R11, RDX # int
106 movl R10, R13 # spill
109 addl R10, #9 # int
10d imull R10, R10 # int
111 sall R10, #1
114 addl R10, R11 # int
117 addl R10, RCX # int
11a addl R10, RBX # int
11d addl R10, R8 # int
120 addl R9, R10 # int
123 addl RBP, R9 # int
126 addl RBP, [RSP + #32 (32-bit)] # int
12a addl R13, #16 # int
12e movl R11, R13 # spill
131 imull R11, R13 # int
135 sall R11, #1
138 cmpl R13, #999999985
13f jl B2 # loop end P=1.000000 C=6554623.000000
We see that there is 1 register that is "spilled" onto the stack.
And for the 2 * i * i
version:
05a B3: # B2 B4 <- B1 B2 Loop: B3-B2 inner main of N18 Freq: 1e+006
05a addl RBX, R11 # int
05d movl [rsp + #32], RBX # spill
061 movl R11, R8 # spill
064 addl R11, #15 # int
068 movl [rsp + #36], R11 # spill
06d movl R11, R8 # spill
070 addl R11, #14 # int
074 movl R10, R9 # spill
077 addl R10, #16 # int
07b movdl XMM2, R10 # spill
080 movl RCX, R9 # spill
083 addl RCX, #14 # int
086 movdl XMM1, RCX # spill
08a movl R10, R9 # spill
08d addl R10, #12 # int
091 movdl XMM4, R10 # spill
096 movl RCX, R9 # spill
099 addl RCX, #10 # int
09c movdl XMM6, RCX # spill
0a0 movl RBX, R9 # spill
0a3 addl RBX, #8 # int
0a6 movl RCX, R9 # spill
0a9 addl RCX, #6 # int
0ac movl RDX, R9 # spill
0af addl RDX, #4 # int
0b2 addl R9, #2 # int
0b6 movl R10, R14 # spill
0b9 addl R10, #22 # int
0bd movdl XMM3, R10 # spill
0c2 movl RDI, R14 # spill
0c5 addl RDI, #20 # int
0c8 movl RAX, R14 # spill
0cb addl RAX, #32 # int
0ce movl RSI, R14 # spill
0d1 addl RSI, #18 # int
0d4 movl R13, R14 # spill
0d7 addl R13, #24 # int
0db movl R10, R14 # spill
0de addl R10, #26 # int
0e2 movl [rsp + #40], R10 # spill
0e7 movl RBP, R14 # spill
0ea addl RBP, #28 # int
0ed imull RBP, R11 # int
0f1 addl R14, #30 # int
0f5 imull R14, [RSP + #36 (32-bit)] # int
0fb movl R10, R8 # spill
0fe addl R10, #11 # int
102 movdl R11, XMM3 # spill
107 imull R11, R10 # int
10b movl [rsp + #44], R11 # spill
110 movl R10, R8 # spill
113 addl R10, #10 # int
117 imull RDI, R10 # int
11b movl R11, R8 # spill
11e addl R11, #8 # int
122 movdl R10, XMM2 # spill
127 imull R10, R11 # int
12b movl [rsp + #48], R10 # spill
130 movl R10, R8 # spill
133 addl R10, #7 # int
137 movdl R11, XMM1 # spill
13c imull R11, R10 # int
140 movl [rsp + #52], R11 # spill
145 movl R11, R8 # spill
148 addl R11, #6 # int
14c movdl R10, XMM4 # spill
151 imull R10, R11 # int
155 movl [rsp + #56], R10 # spill
15a movl R10, R8 # spill
15d addl R10, #5 # int
161 movdl R11, XMM6 # spill
166 imull R11, R10 # int
16a movl [rsp + #60], R11 # spill
16f movl R11, R8 # spill
172 addl R11, #4 # int
176 imull RBX, R11 # int
17a movl R11, R8 # spill
17d addl R11, #3 # int
181 imull RCX, R11 # int
185 movl R10, R8 # spill
188 addl R10, #2 # int
18c imull RDX, R10 # int
190 movl R11, R8 # spill
193 incl R11 # int
196 imull R9, R11 # int
19a addl R9, [RSP + #32 (32-bit)] # int
19f addl R9, RDX # int
1a2 addl R9, RCX # int
1a5 addl R9, RBX # int
1a8 addl R9, [RSP + #60 (32-bit)] # int
1ad addl R9, [RSP + #56 (32-bit)] # int
1b2 addl R9, [RSP + #52 (32-bit)] # int
1b7 addl R9, [RSP + #48 (32-bit)] # int
1bc movl R10, R8 # spill
1bf addl R10, #9 # int
1c3 imull R10, RSI # int
1c7 addl R10, R9 # int
1ca addl R10, RDI # int
1cd addl R10, [RSP + #44 (32-bit)] # int
1d2 movl R11, R8 # spill
1d5 addl R11, #12 # int
1d9 imull R13, R11 # int
1dd addl R13, R10 # int
1e0 movl R10, R8 # spill
1e3 addl R10, #13 # int
1e7 imull R10, [RSP + #40 (32-bit)] # int
1ed addl R10, R13 # int
1f0 addl RBP, R10 # int
1f3 addl R14, RBP # int
1f6 movl R10, R8 # spill
1f9 addl R10, #16 # int
1fd cmpl R10, #999999985
204 jl B2 # loop end P=1.000000 C=7419903.000000
Here we observe much more "spilling" and more accesses to the stack [RSP + ...]
, due to more intermediate results that need to be preserved.
Thus the answer to the question is simple: 2 * (i * i)
is faster than 2 * i * i
because the JIT generates more optimal assembly code for the first case.
But of course it is obvious that neither the first nor the second version is any good; the loop could really benefit from vectorization, since any x86-64 CPU has at least SSE2 support.
So it's an issue of the optimizer; as is often the case, it unrolls too aggressively and shoots itself in the foot, all the while missing out on various other opportunities.
In fact, modern x86-64 CPUs break down the instructions further into micro-ops (µops) and with features like register renaming, µop caches and loop buffers, loop optimization takes a lot more finesse than a simple unrolling for optimal performance. According to Agner Fog's optimization guide:
The gain in performance due to the µop cache can be quite
considerable if the average instruction length is more than 4 bytes.
The following methods of optimizing the use of the µop cache may
be considered:
- Make sure that critical loops are small enough to fit into the µop cache.
- Align the most critical loop entries and function entries by 32.
- Avoid unnecessary loop unrolling.
- Avoid instructions that have extra load time
. . .
Regarding those load times - even the fastest L1D hit costs 4 cycles, an extra register and µop, so yes, even a few accesses to memory will hurt performance in tight loops.
But back to the vectorization opportunity - to see how fast it can be, we can compile a similar C application with GCC, which outright vectorizes it (AVX2 is shown, SSE2 is similar)2:
vmovdqa ymm0, YMMWORD PTR .LC0[rip]
vmovdqa ymm3, YMMWORD PTR .LC1[rip]
xor eax, eax
vpxor xmm2, xmm2, xmm2
.L2:
vpmulld ymm1, ymm0, ymm0
inc eax
vpaddd ymm0, ymm0, ymm3
vpslld ymm1, ymm1, 1
vpaddd ymm2, ymm2, ymm1
cmp eax, 125000000 ; 8 calculations per iteration
jne .L2
vmovdqa xmm0, xmm2
vextracti128 xmm2, ymm2, 1
vpaddd xmm2, xmm0, xmm2
vpsrldq xmm0, xmm2, 8
vpaddd xmm0, xmm2, xmm0
vpsrldq xmm1, xmm0, 4
vpaddd xmm0, xmm0, xmm1
vmovd eax, xmm0
vzeroupper
With run times:
- SSE: 0.24 s, or 2 times faster.
- AVX: 0.15 s, or 3 times faster.
- AVX2: 0.08 s, or 5 times faster.
1To get JIT generated assembly output, get a debug JVM and run with -XX:+PrintOptoAssembly
2The C version is compiled with the -fwrapv
flag, which enables GCC to treat signed integer overflow as a two's-complement wrap-around.
10
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
36
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
12
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just aret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.
– Peter Cordes
Nov 25 at 22:26
8
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole formov
/add-immediate
. e.g.movl RBX, R9
/addl RBX, #8
should beleal ebx, [r9 + 8]
, 1 uop to copy-and-add. Orleal ebx, [r9 + r9 + 16]
to doebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.
– Peter Cordes
Nov 25 at 22:38
6
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
|
show 18 more comments
There is a slight difference in the ordering of the bytecode.
2 * (i * i)
:
iconst_2
iload0
iload0
imul
imul
iadd
vs 2 * i * i
:
iconst_2
iload0
imul
iload0
imul
iadd
At first sight this should not make a difference; if anything the second version is more optimal since it uses one slot less.
So we need to dig deeper into the lower level (JIT)1.
Remember that JIT tends to unroll small loops very aggressively. Indeed we observe a 16x unrolling for the 2 * (i * i)
case:
030 B2: # B2 B3 <- B1 B2 Loop: B2-B2 inner main of N18 Freq: 1e+006
030 addl R11, RBP # int
033 movl RBP, R13 # spill
036 addl RBP, #14 # int
039 imull RBP, RBP # int
03c movl R9, R13 # spill
03f addl R9, #13 # int
043 imull R9, R9 # int
047 sall RBP, #1
049 sall R9, #1
04c movl R8, R13 # spill
04f addl R8, #15 # int
053 movl R10, R8 # spill
056 movdl XMM1, R8 # spill
05b imull R10, R8 # int
05f movl R8, R13 # spill
062 addl R8, #12 # int
066 imull R8, R8 # int
06a sall R10, #1
06d movl [rsp + #32], R10 # spill
072 sall R8, #1
075 movl RBX, R13 # spill
078 addl RBX, #11 # int
07b imull RBX, RBX # int
07e movl RCX, R13 # spill
081 addl RCX, #10 # int
084 imull RCX, RCX # int
087 sall RBX, #1
089 sall RCX, #1
08b movl RDX, R13 # spill
08e addl RDX, #8 # int
091 imull RDX, RDX # int
094 movl RDI, R13 # spill
097 addl RDI, #7 # int
09a imull RDI, RDI # int
09d sall RDX, #1
09f sall RDI, #1
0a1 movl RAX, R13 # spill
0a4 addl RAX, #6 # int
0a7 imull RAX, RAX # int
0aa movl RSI, R13 # spill
0ad addl RSI, #4 # int
0b0 imull RSI, RSI # int
0b3 sall RAX, #1
0b5 sall RSI, #1
0b7 movl R10, R13 # spill
0ba addl R10, #2 # int
0be imull R10, R10 # int
0c2 movl R14, R13 # spill
0c5 incl R14 # int
0c8 imull R14, R14 # int
0cc sall R10, #1
0cf sall R14, #1
0d2 addl R14, R11 # int
0d5 addl R14, R10 # int
0d8 movl R10, R13 # spill
0db addl R10, #3 # int
0df imull R10, R10 # int
0e3 movl R11, R13 # spill
0e6 addl R11, #5 # int
0ea imull R11, R11 # int
0ee sall R10, #1
0f1 addl R10, R14 # int
0f4 addl R10, RSI # int
0f7 sall R11, #1
0fa addl R11, R10 # int
0fd addl R11, RAX # int
100 addl R11, RDI # int
103 addl R11, RDX # int
106 movl R10, R13 # spill
109 addl R10, #9 # int
10d imull R10, R10 # int
111 sall R10, #1
114 addl R10, R11 # int
117 addl R10, RCX # int
11a addl R10, RBX # int
11d addl R10, R8 # int
120 addl R9, R10 # int
123 addl RBP, R9 # int
126 addl RBP, [RSP + #32 (32-bit)] # int
12a addl R13, #16 # int
12e movl R11, R13 # spill
131 imull R11, R13 # int
135 sall R11, #1
138 cmpl R13, #999999985
13f jl B2 # loop end P=1.000000 C=6554623.000000
We see that there is 1 register that is "spilled" onto the stack.
And for the 2 * i * i
version:
05a B3: # B2 B4 <- B1 B2 Loop: B3-B2 inner main of N18 Freq: 1e+006
05a addl RBX, R11 # int
05d movl [rsp + #32], RBX # spill
061 movl R11, R8 # spill
064 addl R11, #15 # int
068 movl [rsp + #36], R11 # spill
06d movl R11, R8 # spill
070 addl R11, #14 # int
074 movl R10, R9 # spill
077 addl R10, #16 # int
07b movdl XMM2, R10 # spill
080 movl RCX, R9 # spill
083 addl RCX, #14 # int
086 movdl XMM1, RCX # spill
08a movl R10, R9 # spill
08d addl R10, #12 # int
091 movdl XMM4, R10 # spill
096 movl RCX, R9 # spill
099 addl RCX, #10 # int
09c movdl XMM6, RCX # spill
0a0 movl RBX, R9 # spill
0a3 addl RBX, #8 # int
0a6 movl RCX, R9 # spill
0a9 addl RCX, #6 # int
0ac movl RDX, R9 # spill
0af addl RDX, #4 # int
0b2 addl R9, #2 # int
0b6 movl R10, R14 # spill
0b9 addl R10, #22 # int
0bd movdl XMM3, R10 # spill
0c2 movl RDI, R14 # spill
0c5 addl RDI, #20 # int
0c8 movl RAX, R14 # spill
0cb addl RAX, #32 # int
0ce movl RSI, R14 # spill
0d1 addl RSI, #18 # int
0d4 movl R13, R14 # spill
0d7 addl R13, #24 # int
0db movl R10, R14 # spill
0de addl R10, #26 # int
0e2 movl [rsp + #40], R10 # spill
0e7 movl RBP, R14 # spill
0ea addl RBP, #28 # int
0ed imull RBP, R11 # int
0f1 addl R14, #30 # int
0f5 imull R14, [RSP + #36 (32-bit)] # int
0fb movl R10, R8 # spill
0fe addl R10, #11 # int
102 movdl R11, XMM3 # spill
107 imull R11, R10 # int
10b movl [rsp + #44], R11 # spill
110 movl R10, R8 # spill
113 addl R10, #10 # int
117 imull RDI, R10 # int
11b movl R11, R8 # spill
11e addl R11, #8 # int
122 movdl R10, XMM2 # spill
127 imull R10, R11 # int
12b movl [rsp + #48], R10 # spill
130 movl R10, R8 # spill
133 addl R10, #7 # int
137 movdl R11, XMM1 # spill
13c imull R11, R10 # int
140 movl [rsp + #52], R11 # spill
145 movl R11, R8 # spill
148 addl R11, #6 # int
14c movdl R10, XMM4 # spill
151 imull R10, R11 # int
155 movl [rsp + #56], R10 # spill
15a movl R10, R8 # spill
15d addl R10, #5 # int
161 movdl R11, XMM6 # spill
166 imull R11, R10 # int
16a movl [rsp + #60], R11 # spill
16f movl R11, R8 # spill
172 addl R11, #4 # int
176 imull RBX, R11 # int
17a movl R11, R8 # spill
17d addl R11, #3 # int
181 imull RCX, R11 # int
185 movl R10, R8 # spill
188 addl R10, #2 # int
18c imull RDX, R10 # int
190 movl R11, R8 # spill
193 incl R11 # int
196 imull R9, R11 # int
19a addl R9, [RSP + #32 (32-bit)] # int
19f addl R9, RDX # int
1a2 addl R9, RCX # int
1a5 addl R9, RBX # int
1a8 addl R9, [RSP + #60 (32-bit)] # int
1ad addl R9, [RSP + #56 (32-bit)] # int
1b2 addl R9, [RSP + #52 (32-bit)] # int
1b7 addl R9, [RSP + #48 (32-bit)] # int
1bc movl R10, R8 # spill
1bf addl R10, #9 # int
1c3 imull R10, RSI # int
1c7 addl R10, R9 # int
1ca addl R10, RDI # int
1cd addl R10, [RSP + #44 (32-bit)] # int
1d2 movl R11, R8 # spill
1d5 addl R11, #12 # int
1d9 imull R13, R11 # int
1dd addl R13, R10 # int
1e0 movl R10, R8 # spill
1e3 addl R10, #13 # int
1e7 imull R10, [RSP + #40 (32-bit)] # int
1ed addl R10, R13 # int
1f0 addl RBP, R10 # int
1f3 addl R14, RBP # int
1f6 movl R10, R8 # spill
1f9 addl R10, #16 # int
1fd cmpl R10, #999999985
204 jl B2 # loop end P=1.000000 C=7419903.000000
Here we observe much more "spilling" and more accesses to the stack [RSP + ...]
, due to more intermediate results that need to be preserved.
Thus the answer to the question is simple: 2 * (i * i)
is faster than 2 * i * i
because the JIT generates more optimal assembly code for the first case.
But of course it is obvious that neither the first nor the second version is any good; the loop could really benefit from vectorization, since any x86-64 CPU has at least SSE2 support.
So it's an issue of the optimizer; as is often the case, it unrolls too aggressively and shoots itself in the foot, all the while missing out on various other opportunities.
In fact, modern x86-64 CPUs break down the instructions further into micro-ops (µops) and with features like register renaming, µop caches and loop buffers, loop optimization takes a lot more finesse than a simple unrolling for optimal performance. According to Agner Fog's optimization guide:
The gain in performance due to the µop cache can be quite
considerable if the average instruction length is more than 4 bytes.
The following methods of optimizing the use of the µop cache may
be considered:
- Make sure that critical loops are small enough to fit into the µop cache.
- Align the most critical loop entries and function entries by 32.
- Avoid unnecessary loop unrolling.
- Avoid instructions that have extra load time
. . .
Regarding those load times - even the fastest L1D hit costs 4 cycles, an extra register and µop, so yes, even a few accesses to memory will hurt performance in tight loops.
But back to the vectorization opportunity - to see how fast it can be, we can compile a similar C application with GCC, which outright vectorizes it (AVX2 is shown, SSE2 is similar)2:
vmovdqa ymm0, YMMWORD PTR .LC0[rip]
vmovdqa ymm3, YMMWORD PTR .LC1[rip]
xor eax, eax
vpxor xmm2, xmm2, xmm2
.L2:
vpmulld ymm1, ymm0, ymm0
inc eax
vpaddd ymm0, ymm0, ymm3
vpslld ymm1, ymm1, 1
vpaddd ymm2, ymm2, ymm1
cmp eax, 125000000 ; 8 calculations per iteration
jne .L2
vmovdqa xmm0, xmm2
vextracti128 xmm2, ymm2, 1
vpaddd xmm2, xmm0, xmm2
vpsrldq xmm0, xmm2, 8
vpaddd xmm0, xmm2, xmm0
vpsrldq xmm1, xmm0, 4
vpaddd xmm0, xmm0, xmm1
vmovd eax, xmm0
vzeroupper
With run times:
- SSE: 0.24 s, or 2 times faster.
- AVX: 0.15 s, or 3 times faster.
- AVX2: 0.08 s, or 5 times faster.
1To get JIT generated assembly output, get a debug JVM and run with -XX:+PrintOptoAssembly
2The C version is compiled with the -fwrapv
flag, which enables GCC to treat signed integer overflow as a two's-complement wrap-around.
10
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
36
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
12
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just aret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.
– Peter Cordes
Nov 25 at 22:26
8
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole formov
/add-immediate
. e.g.movl RBX, R9
/addl RBX, #8
should beleal ebx, [r9 + 8]
, 1 uop to copy-and-add. Orleal ebx, [r9 + r9 + 16]
to doebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.
– Peter Cordes
Nov 25 at 22:38
6
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
|
show 18 more comments
There is a slight difference in the ordering of the bytecode.
2 * (i * i)
:
iconst_2
iload0
iload0
imul
imul
iadd
vs 2 * i * i
:
iconst_2
iload0
imul
iload0
imul
iadd
At first sight this should not make a difference; if anything the second version is more optimal since it uses one slot less.
So we need to dig deeper into the lower level (JIT)1.
Remember that JIT tends to unroll small loops very aggressively. Indeed we observe a 16x unrolling for the 2 * (i * i)
case:
030 B2: # B2 B3 <- B1 B2 Loop: B2-B2 inner main of N18 Freq: 1e+006
030 addl R11, RBP # int
033 movl RBP, R13 # spill
036 addl RBP, #14 # int
039 imull RBP, RBP # int
03c movl R9, R13 # spill
03f addl R9, #13 # int
043 imull R9, R9 # int
047 sall RBP, #1
049 sall R9, #1
04c movl R8, R13 # spill
04f addl R8, #15 # int
053 movl R10, R8 # spill
056 movdl XMM1, R8 # spill
05b imull R10, R8 # int
05f movl R8, R13 # spill
062 addl R8, #12 # int
066 imull R8, R8 # int
06a sall R10, #1
06d movl [rsp + #32], R10 # spill
072 sall R8, #1
075 movl RBX, R13 # spill
078 addl RBX, #11 # int
07b imull RBX, RBX # int
07e movl RCX, R13 # spill
081 addl RCX, #10 # int
084 imull RCX, RCX # int
087 sall RBX, #1
089 sall RCX, #1
08b movl RDX, R13 # spill
08e addl RDX, #8 # int
091 imull RDX, RDX # int
094 movl RDI, R13 # spill
097 addl RDI, #7 # int
09a imull RDI, RDI # int
09d sall RDX, #1
09f sall RDI, #1
0a1 movl RAX, R13 # spill
0a4 addl RAX, #6 # int
0a7 imull RAX, RAX # int
0aa movl RSI, R13 # spill
0ad addl RSI, #4 # int
0b0 imull RSI, RSI # int
0b3 sall RAX, #1
0b5 sall RSI, #1
0b7 movl R10, R13 # spill
0ba addl R10, #2 # int
0be imull R10, R10 # int
0c2 movl R14, R13 # spill
0c5 incl R14 # int
0c8 imull R14, R14 # int
0cc sall R10, #1
0cf sall R14, #1
0d2 addl R14, R11 # int
0d5 addl R14, R10 # int
0d8 movl R10, R13 # spill
0db addl R10, #3 # int
0df imull R10, R10 # int
0e3 movl R11, R13 # spill
0e6 addl R11, #5 # int
0ea imull R11, R11 # int
0ee sall R10, #1
0f1 addl R10, R14 # int
0f4 addl R10, RSI # int
0f7 sall R11, #1
0fa addl R11, R10 # int
0fd addl R11, RAX # int
100 addl R11, RDI # int
103 addl R11, RDX # int
106 movl R10, R13 # spill
109 addl R10, #9 # int
10d imull R10, R10 # int
111 sall R10, #1
114 addl R10, R11 # int
117 addl R10, RCX # int
11a addl R10, RBX # int
11d addl R10, R8 # int
120 addl R9, R10 # int
123 addl RBP, R9 # int
126 addl RBP, [RSP + #32 (32-bit)] # int
12a addl R13, #16 # int
12e movl R11, R13 # spill
131 imull R11, R13 # int
135 sall R11, #1
138 cmpl R13, #999999985
13f jl B2 # loop end P=1.000000 C=6554623.000000
We see that there is 1 register that is "spilled" onto the stack.
And for the 2 * i * i
version:
05a B3: # B2 B4 <- B1 B2 Loop: B3-B2 inner main of N18 Freq: 1e+006
05a addl RBX, R11 # int
05d movl [rsp + #32], RBX # spill
061 movl R11, R8 # spill
064 addl R11, #15 # int
068 movl [rsp + #36], R11 # spill
06d movl R11, R8 # spill
070 addl R11, #14 # int
074 movl R10, R9 # spill
077 addl R10, #16 # int
07b movdl XMM2, R10 # spill
080 movl RCX, R9 # spill
083 addl RCX, #14 # int
086 movdl XMM1, RCX # spill
08a movl R10, R9 # spill
08d addl R10, #12 # int
091 movdl XMM4, R10 # spill
096 movl RCX, R9 # spill
099 addl RCX, #10 # int
09c movdl XMM6, RCX # spill
0a0 movl RBX, R9 # spill
0a3 addl RBX, #8 # int
0a6 movl RCX, R9 # spill
0a9 addl RCX, #6 # int
0ac movl RDX, R9 # spill
0af addl RDX, #4 # int
0b2 addl R9, #2 # int
0b6 movl R10, R14 # spill
0b9 addl R10, #22 # int
0bd movdl XMM3, R10 # spill
0c2 movl RDI, R14 # spill
0c5 addl RDI, #20 # int
0c8 movl RAX, R14 # spill
0cb addl RAX, #32 # int
0ce movl RSI, R14 # spill
0d1 addl RSI, #18 # int
0d4 movl R13, R14 # spill
0d7 addl R13, #24 # int
0db movl R10, R14 # spill
0de addl R10, #26 # int
0e2 movl [rsp + #40], R10 # spill
0e7 movl RBP, R14 # spill
0ea addl RBP, #28 # int
0ed imull RBP, R11 # int
0f1 addl R14, #30 # int
0f5 imull R14, [RSP + #36 (32-bit)] # int
0fb movl R10, R8 # spill
0fe addl R10, #11 # int
102 movdl R11, XMM3 # spill
107 imull R11, R10 # int
10b movl [rsp + #44], R11 # spill
110 movl R10, R8 # spill
113 addl R10, #10 # int
117 imull RDI, R10 # int
11b movl R11, R8 # spill
11e addl R11, #8 # int
122 movdl R10, XMM2 # spill
127 imull R10, R11 # int
12b movl [rsp + #48], R10 # spill
130 movl R10, R8 # spill
133 addl R10, #7 # int
137 movdl R11, XMM1 # spill
13c imull R11, R10 # int
140 movl [rsp + #52], R11 # spill
145 movl R11, R8 # spill
148 addl R11, #6 # int
14c movdl R10, XMM4 # spill
151 imull R10, R11 # int
155 movl [rsp + #56], R10 # spill
15a movl R10, R8 # spill
15d addl R10, #5 # int
161 movdl R11, XMM6 # spill
166 imull R11, R10 # int
16a movl [rsp + #60], R11 # spill
16f movl R11, R8 # spill
172 addl R11, #4 # int
176 imull RBX, R11 # int
17a movl R11, R8 # spill
17d addl R11, #3 # int
181 imull RCX, R11 # int
185 movl R10, R8 # spill
188 addl R10, #2 # int
18c imull RDX, R10 # int
190 movl R11, R8 # spill
193 incl R11 # int
196 imull R9, R11 # int
19a addl R9, [RSP + #32 (32-bit)] # int
19f addl R9, RDX # int
1a2 addl R9, RCX # int
1a5 addl R9, RBX # int
1a8 addl R9, [RSP + #60 (32-bit)] # int
1ad addl R9, [RSP + #56 (32-bit)] # int
1b2 addl R9, [RSP + #52 (32-bit)] # int
1b7 addl R9, [RSP + #48 (32-bit)] # int
1bc movl R10, R8 # spill
1bf addl R10, #9 # int
1c3 imull R10, RSI # int
1c7 addl R10, R9 # int
1ca addl R10, RDI # int
1cd addl R10, [RSP + #44 (32-bit)] # int
1d2 movl R11, R8 # spill
1d5 addl R11, #12 # int
1d9 imull R13, R11 # int
1dd addl R13, R10 # int
1e0 movl R10, R8 # spill
1e3 addl R10, #13 # int
1e7 imull R10, [RSP + #40 (32-bit)] # int
1ed addl R10, R13 # int
1f0 addl RBP, R10 # int
1f3 addl R14, RBP # int
1f6 movl R10, R8 # spill
1f9 addl R10, #16 # int
1fd cmpl R10, #999999985
204 jl B2 # loop end P=1.000000 C=7419903.000000
Here we observe much more "spilling" and more accesses to the stack [RSP + ...]
, due to more intermediate results that need to be preserved.
Thus the answer to the question is simple: 2 * (i * i)
is faster than 2 * i * i
because the JIT generates more optimal assembly code for the first case.
But of course it is obvious that neither the first nor the second version is any good; the loop could really benefit from vectorization, since any x86-64 CPU has at least SSE2 support.
So it's an issue of the optimizer; as is often the case, it unrolls too aggressively and shoots itself in the foot, all the while missing out on various other opportunities.
In fact, modern x86-64 CPUs break down the instructions further into micro-ops (µops) and with features like register renaming, µop caches and loop buffers, loop optimization takes a lot more finesse than a simple unrolling for optimal performance. According to Agner Fog's optimization guide:
The gain in performance due to the µop cache can be quite
considerable if the average instruction length is more than 4 bytes.
The following methods of optimizing the use of the µop cache may
be considered:
- Make sure that critical loops are small enough to fit into the µop cache.
- Align the most critical loop entries and function entries by 32.
- Avoid unnecessary loop unrolling.
- Avoid instructions that have extra load time
. . .
Regarding those load times - even the fastest L1D hit costs 4 cycles, an extra register and µop, so yes, even a few accesses to memory will hurt performance in tight loops.
But back to the vectorization opportunity - to see how fast it can be, we can compile a similar C application with GCC, which outright vectorizes it (AVX2 is shown, SSE2 is similar)2:
vmovdqa ymm0, YMMWORD PTR .LC0[rip]
vmovdqa ymm3, YMMWORD PTR .LC1[rip]
xor eax, eax
vpxor xmm2, xmm2, xmm2
.L2:
vpmulld ymm1, ymm0, ymm0
inc eax
vpaddd ymm0, ymm0, ymm3
vpslld ymm1, ymm1, 1
vpaddd ymm2, ymm2, ymm1
cmp eax, 125000000 ; 8 calculations per iteration
jne .L2
vmovdqa xmm0, xmm2
vextracti128 xmm2, ymm2, 1
vpaddd xmm2, xmm0, xmm2
vpsrldq xmm0, xmm2, 8
vpaddd xmm0, xmm2, xmm0
vpsrldq xmm1, xmm0, 4
vpaddd xmm0, xmm0, xmm1
vmovd eax, xmm0
vzeroupper
With run times:
- SSE: 0.24 s, or 2 times faster.
- AVX: 0.15 s, or 3 times faster.
- AVX2: 0.08 s, or 5 times faster.
1To get JIT generated assembly output, get a debug JVM and run with -XX:+PrintOptoAssembly
2The C version is compiled with the -fwrapv
flag, which enables GCC to treat signed integer overflow as a two's-complement wrap-around.
There is a slight difference in the ordering of the bytecode.
2 * (i * i)
:
iconst_2
iload0
iload0
imul
imul
iadd
vs 2 * i * i
:
iconst_2
iload0
imul
iload0
imul
iadd
At first sight this should not make a difference; if anything the second version is more optimal since it uses one slot less.
So we need to dig deeper into the lower level (JIT)1.
Remember that JIT tends to unroll small loops very aggressively. Indeed we observe a 16x unrolling for the 2 * (i * i)
case:
030 B2: # B2 B3 <- B1 B2 Loop: B2-B2 inner main of N18 Freq: 1e+006
030 addl R11, RBP # int
033 movl RBP, R13 # spill
036 addl RBP, #14 # int
039 imull RBP, RBP # int
03c movl R9, R13 # spill
03f addl R9, #13 # int
043 imull R9, R9 # int
047 sall RBP, #1
049 sall R9, #1
04c movl R8, R13 # spill
04f addl R8, #15 # int
053 movl R10, R8 # spill
056 movdl XMM1, R8 # spill
05b imull R10, R8 # int
05f movl R8, R13 # spill
062 addl R8, #12 # int
066 imull R8, R8 # int
06a sall R10, #1
06d movl [rsp + #32], R10 # spill
072 sall R8, #1
075 movl RBX, R13 # spill
078 addl RBX, #11 # int
07b imull RBX, RBX # int
07e movl RCX, R13 # spill
081 addl RCX, #10 # int
084 imull RCX, RCX # int
087 sall RBX, #1
089 sall RCX, #1
08b movl RDX, R13 # spill
08e addl RDX, #8 # int
091 imull RDX, RDX # int
094 movl RDI, R13 # spill
097 addl RDI, #7 # int
09a imull RDI, RDI # int
09d sall RDX, #1
09f sall RDI, #1
0a1 movl RAX, R13 # spill
0a4 addl RAX, #6 # int
0a7 imull RAX, RAX # int
0aa movl RSI, R13 # spill
0ad addl RSI, #4 # int
0b0 imull RSI, RSI # int
0b3 sall RAX, #1
0b5 sall RSI, #1
0b7 movl R10, R13 # spill
0ba addl R10, #2 # int
0be imull R10, R10 # int
0c2 movl R14, R13 # spill
0c5 incl R14 # int
0c8 imull R14, R14 # int
0cc sall R10, #1
0cf sall R14, #1
0d2 addl R14, R11 # int
0d5 addl R14, R10 # int
0d8 movl R10, R13 # spill
0db addl R10, #3 # int
0df imull R10, R10 # int
0e3 movl R11, R13 # spill
0e6 addl R11, #5 # int
0ea imull R11, R11 # int
0ee sall R10, #1
0f1 addl R10, R14 # int
0f4 addl R10, RSI # int
0f7 sall R11, #1
0fa addl R11, R10 # int
0fd addl R11, RAX # int
100 addl R11, RDI # int
103 addl R11, RDX # int
106 movl R10, R13 # spill
109 addl R10, #9 # int
10d imull R10, R10 # int
111 sall R10, #1
114 addl R10, R11 # int
117 addl R10, RCX # int
11a addl R10, RBX # int
11d addl R10, R8 # int
120 addl R9, R10 # int
123 addl RBP, R9 # int
126 addl RBP, [RSP + #32 (32-bit)] # int
12a addl R13, #16 # int
12e movl R11, R13 # spill
131 imull R11, R13 # int
135 sall R11, #1
138 cmpl R13, #999999985
13f jl B2 # loop end P=1.000000 C=6554623.000000
We see that there is 1 register that is "spilled" onto the stack.
And for the 2 * i * i
version:
05a B3: # B2 B4 <- B1 B2 Loop: B3-B2 inner main of N18 Freq: 1e+006
05a addl RBX, R11 # int
05d movl [rsp + #32], RBX # spill
061 movl R11, R8 # spill
064 addl R11, #15 # int
068 movl [rsp + #36], R11 # spill
06d movl R11, R8 # spill
070 addl R11, #14 # int
074 movl R10, R9 # spill
077 addl R10, #16 # int
07b movdl XMM2, R10 # spill
080 movl RCX, R9 # spill
083 addl RCX, #14 # int
086 movdl XMM1, RCX # spill
08a movl R10, R9 # spill
08d addl R10, #12 # int
091 movdl XMM4, R10 # spill
096 movl RCX, R9 # spill
099 addl RCX, #10 # int
09c movdl XMM6, RCX # spill
0a0 movl RBX, R9 # spill
0a3 addl RBX, #8 # int
0a6 movl RCX, R9 # spill
0a9 addl RCX, #6 # int
0ac movl RDX, R9 # spill
0af addl RDX, #4 # int
0b2 addl R9, #2 # int
0b6 movl R10, R14 # spill
0b9 addl R10, #22 # int
0bd movdl XMM3, R10 # spill
0c2 movl RDI, R14 # spill
0c5 addl RDI, #20 # int
0c8 movl RAX, R14 # spill
0cb addl RAX, #32 # int
0ce movl RSI, R14 # spill
0d1 addl RSI, #18 # int
0d4 movl R13, R14 # spill
0d7 addl R13, #24 # int
0db movl R10, R14 # spill
0de addl R10, #26 # int
0e2 movl [rsp + #40], R10 # spill
0e7 movl RBP, R14 # spill
0ea addl RBP, #28 # int
0ed imull RBP, R11 # int
0f1 addl R14, #30 # int
0f5 imull R14, [RSP + #36 (32-bit)] # int
0fb movl R10, R8 # spill
0fe addl R10, #11 # int
102 movdl R11, XMM3 # spill
107 imull R11, R10 # int
10b movl [rsp + #44], R11 # spill
110 movl R10, R8 # spill
113 addl R10, #10 # int
117 imull RDI, R10 # int
11b movl R11, R8 # spill
11e addl R11, #8 # int
122 movdl R10, XMM2 # spill
127 imull R10, R11 # int
12b movl [rsp + #48], R10 # spill
130 movl R10, R8 # spill
133 addl R10, #7 # int
137 movdl R11, XMM1 # spill
13c imull R11, R10 # int
140 movl [rsp + #52], R11 # spill
145 movl R11, R8 # spill
148 addl R11, #6 # int
14c movdl R10, XMM4 # spill
151 imull R10, R11 # int
155 movl [rsp + #56], R10 # spill
15a movl R10, R8 # spill
15d addl R10, #5 # int
161 movdl R11, XMM6 # spill
166 imull R11, R10 # int
16a movl [rsp + #60], R11 # spill
16f movl R11, R8 # spill
172 addl R11, #4 # int
176 imull RBX, R11 # int
17a movl R11, R8 # spill
17d addl R11, #3 # int
181 imull RCX, R11 # int
185 movl R10, R8 # spill
188 addl R10, #2 # int
18c imull RDX, R10 # int
190 movl R11, R8 # spill
193 incl R11 # int
196 imull R9, R11 # int
19a addl R9, [RSP + #32 (32-bit)] # int
19f addl R9, RDX # int
1a2 addl R9, RCX # int
1a5 addl R9, RBX # int
1a8 addl R9, [RSP + #60 (32-bit)] # int
1ad addl R9, [RSP + #56 (32-bit)] # int
1b2 addl R9, [RSP + #52 (32-bit)] # int
1b7 addl R9, [RSP + #48 (32-bit)] # int
1bc movl R10, R8 # spill
1bf addl R10, #9 # int
1c3 imull R10, RSI # int
1c7 addl R10, R9 # int
1ca addl R10, RDI # int
1cd addl R10, [RSP + #44 (32-bit)] # int
1d2 movl R11, R8 # spill
1d5 addl R11, #12 # int
1d9 imull R13, R11 # int
1dd addl R13, R10 # int
1e0 movl R10, R8 # spill
1e3 addl R10, #13 # int
1e7 imull R10, [RSP + #40 (32-bit)] # int
1ed addl R10, R13 # int
1f0 addl RBP, R10 # int
1f3 addl R14, RBP # int
1f6 movl R10, R8 # spill
1f9 addl R10, #16 # int
1fd cmpl R10, #999999985
204 jl B2 # loop end P=1.000000 C=7419903.000000
Here we observe much more "spilling" and more accesses to the stack [RSP + ...]
, due to more intermediate results that need to be preserved.
Thus the answer to the question is simple: 2 * (i * i)
is faster than 2 * i * i
because the JIT generates more optimal assembly code for the first case.
But of course it is obvious that neither the first nor the second version is any good; the loop could really benefit from vectorization, since any x86-64 CPU has at least SSE2 support.
So it's an issue of the optimizer; as is often the case, it unrolls too aggressively and shoots itself in the foot, all the while missing out on various other opportunities.
In fact, modern x86-64 CPUs break down the instructions further into micro-ops (µops) and with features like register renaming, µop caches and loop buffers, loop optimization takes a lot more finesse than a simple unrolling for optimal performance. According to Agner Fog's optimization guide:
The gain in performance due to the µop cache can be quite
considerable if the average instruction length is more than 4 bytes.
The following methods of optimizing the use of the µop cache may
be considered:
- Make sure that critical loops are small enough to fit into the µop cache.
- Align the most critical loop entries and function entries by 32.
- Avoid unnecessary loop unrolling.
- Avoid instructions that have extra load time
. . .
Regarding those load times - even the fastest L1D hit costs 4 cycles, an extra register and µop, so yes, even a few accesses to memory will hurt performance in tight loops.
But back to the vectorization opportunity - to see how fast it can be, we can compile a similar C application with GCC, which outright vectorizes it (AVX2 is shown, SSE2 is similar)2:
vmovdqa ymm0, YMMWORD PTR .LC0[rip]
vmovdqa ymm3, YMMWORD PTR .LC1[rip]
xor eax, eax
vpxor xmm2, xmm2, xmm2
.L2:
vpmulld ymm1, ymm0, ymm0
inc eax
vpaddd ymm0, ymm0, ymm3
vpslld ymm1, ymm1, 1
vpaddd ymm2, ymm2, ymm1
cmp eax, 125000000 ; 8 calculations per iteration
jne .L2
vmovdqa xmm0, xmm2
vextracti128 xmm2, ymm2, 1
vpaddd xmm2, xmm0, xmm2
vpsrldq xmm0, xmm2, 8
vpaddd xmm0, xmm2, xmm0
vpsrldq xmm1, xmm0, 4
vpaddd xmm0, xmm0, xmm1
vmovd eax, xmm0
vzeroupper
With run times:
- SSE: 0.24 s, or 2 times faster.
- AVX: 0.15 s, or 3 times faster.
- AVX2: 0.08 s, or 5 times faster.
1To get JIT generated assembly output, get a debug JVM and run with -XX:+PrintOptoAssembly
2The C version is compiled with the -fwrapv
flag, which enables GCC to treat signed integer overflow as a two's-complement wrap-around.
edited Nov 27 at 9:18
Ian Kemp
16.4k126797
16.4k126797
answered Nov 23 at 22:40
rustyx
28.7k696136
28.7k696136
10
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
36
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
12
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just aret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.
– Peter Cordes
Nov 25 at 22:26
8
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole formov
/add-immediate
. e.g.movl RBX, R9
/addl RBX, #8
should beleal ebx, [r9 + 8]
, 1 uop to copy-and-add. Orleal ebx, [r9 + r9 + 16]
to doebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.
– Peter Cordes
Nov 25 at 22:38
6
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
|
show 18 more comments
10
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
36
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
12
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just aret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.
– Peter Cordes
Nov 25 at 22:26
8
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole formov
/add-immediate
. e.g.movl RBX, R9
/addl RBX, #8
should beleal ebx, [r9 + 8]
, 1 uop to copy-and-add. Orleal ebx, [r9 + r9 + 16]
to doebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.
– Peter Cordes
Nov 25 at 22:38
6
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
10
10
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
The single biggest problem the optimizer encounters in the C example is the undefined behavior invoked by signed integer overflow. Which, otherwise, would probably result in simply loading a constant as the whole loop can be calculated at compiletime.
– Damon
Nov 25 at 18:28
36
36
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
@Damon Why would undefined behavior be a problem for the optimizer? If the optimizer sees it overflows when trying to calculate the result, that just means it can optimize it however it wants, because the behavior is undefined.
– Runemoro
Nov 25 at 18:51
12
12
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just a
ret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.– Peter Cordes
Nov 25 at 22:26
@Runemoro: if the optimizer proves that calling the function will inevitably result in undefined behaviour, it could choose to assume that the function will never be called, and emit no body for it. Or emit just a
ret
instruction, or emit a label and no ret instruction so execution just falls through. GCC does in fact behave this was sometimes when it encounters UB, though. For example: why ret disappear with optimization?. You definitely want to compile well-formed code to be sure the asm is sane.– Peter Cordes
Nov 25 at 22:26
8
8
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole for
mov
/ add-immediate
. e.g. movl RBX, R9
/ addl RBX, #8
should be leal ebx, [r9 + 8]
, 1 uop to copy-and-add. Or leal ebx, [r9 + r9 + 16]
to do ebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.– Peter Cordes
Nov 25 at 22:38
It's probably just a front-end uop throughput bottleneck because of the inefficient code-gen. It's not even using LEA as a peephole for
mov
/ add-immediate
. e.g. movl RBX, R9
/ addl RBX, #8
should be leal ebx, [r9 + 8]
, 1 uop to copy-and-add. Or leal ebx, [r9 + r9 + 16]
to do ebx = 2*(r9+8)
. So yeah, unrolling to the point of spilling is dumb, and so is naive braindead codegen that doesn't take advantage of integer identities and associative integer math.– Peter Cordes
Nov 25 at 22:38
6
6
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
Vectorization for sequential reduction was disabled in C2 (bugs.openjdk.java.net/browse/JDK-8078563), but is now being considered for re-enabling (bugs.openjdk.java.net/browse/JDK-8188313).
– pron
Nov 30 at 14:31
|
show 18 more comments
When the multiplication is 2 * (i * i)
, the JVM is able to factor out the multiplication by 2
from the loop, resulting in this equivalent but more efficient code:
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
but when the multiplication is (2 * i) * i
, the JVM doesn't optimize it since the multiplication by a constant is no longer right before the addition.
Here are a few reasons why I think this is the case:
- Adding an
if (n == 0) n = 1
statement at the start of the loop results in both versions being as efficient, since factoring out the multiplication no longer guarantees that the result will be the same - The optimized version (by factoring out the multiplication by 2) is exactly as fast as the
2 * (i * i)
version
Here is the test code that I used to draw these conclusions:
public static void main(String args) {
long fastVersion = 0;
long slowVersion = 0;
long optimizedVersion = 0;
long modifiedFastVersion = 0;
long modifiedSlowVersion = 0;
for (int i = 0; i < 10; i++) {
fastVersion += fastVersion();
slowVersion += slowVersion();
optimizedVersion += optimizedVersion();
modifiedFastVersion += modifiedFastVersion();
modifiedSlowVersion += modifiedSlowVersion();
}
System.out.println("Fast version: " + (double) fastVersion / 1000000000 + " s");
System.out.println("Slow version: " + (double) slowVersion / 1000000000 + " s");
System.out.println("Optimized version: " + (double) optimizedVersion / 1000000000 + " s");
System.out.println("Modified fast version: " + (double) modifiedFastVersion / 1000000000 + " s");
System.out.println("Modified slow version: " + (double) modifiedSlowVersion / 1000000000 + " s");
}
private static long fastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long slowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
private static long optimizedVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
return System.nanoTime() - startTime;
}
private static long modifiedFastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long modifiedSlowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
And here are the results:
Fast version: 5.7274411 s
Slow version: 7.6190804 s
Optimized version: 5.1348007 s
Modified fast version: 7.1492705 s
Modified slow version: 7.2952668 s
3
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
2
I think on the optimizedVersion, it should ben *= 2000000000;
– StefansArya
Nov 24 at 1:19
4
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.
– Martin Bonner
Nov 26 at 15:57
5
The math equation was just like this2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.
– StefansArya
Nov 26 at 17:22
2
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
|
show 2 more comments
When the multiplication is 2 * (i * i)
, the JVM is able to factor out the multiplication by 2
from the loop, resulting in this equivalent but more efficient code:
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
but when the multiplication is (2 * i) * i
, the JVM doesn't optimize it since the multiplication by a constant is no longer right before the addition.
Here are a few reasons why I think this is the case:
- Adding an
if (n == 0) n = 1
statement at the start of the loop results in both versions being as efficient, since factoring out the multiplication no longer guarantees that the result will be the same - The optimized version (by factoring out the multiplication by 2) is exactly as fast as the
2 * (i * i)
version
Here is the test code that I used to draw these conclusions:
public static void main(String args) {
long fastVersion = 0;
long slowVersion = 0;
long optimizedVersion = 0;
long modifiedFastVersion = 0;
long modifiedSlowVersion = 0;
for (int i = 0; i < 10; i++) {
fastVersion += fastVersion();
slowVersion += slowVersion();
optimizedVersion += optimizedVersion();
modifiedFastVersion += modifiedFastVersion();
modifiedSlowVersion += modifiedSlowVersion();
}
System.out.println("Fast version: " + (double) fastVersion / 1000000000 + " s");
System.out.println("Slow version: " + (double) slowVersion / 1000000000 + " s");
System.out.println("Optimized version: " + (double) optimizedVersion / 1000000000 + " s");
System.out.println("Modified fast version: " + (double) modifiedFastVersion / 1000000000 + " s");
System.out.println("Modified slow version: " + (double) modifiedSlowVersion / 1000000000 + " s");
}
private static long fastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long slowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
private static long optimizedVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
return System.nanoTime() - startTime;
}
private static long modifiedFastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long modifiedSlowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
And here are the results:
Fast version: 5.7274411 s
Slow version: 7.6190804 s
Optimized version: 5.1348007 s
Modified fast version: 7.1492705 s
Modified slow version: 7.2952668 s
3
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
2
I think on the optimizedVersion, it should ben *= 2000000000;
– StefansArya
Nov 24 at 1:19
4
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.
– Martin Bonner
Nov 26 at 15:57
5
The math equation was just like this2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.
– StefansArya
Nov 26 at 17:22
2
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
|
show 2 more comments
When the multiplication is 2 * (i * i)
, the JVM is able to factor out the multiplication by 2
from the loop, resulting in this equivalent but more efficient code:
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
but when the multiplication is (2 * i) * i
, the JVM doesn't optimize it since the multiplication by a constant is no longer right before the addition.
Here are a few reasons why I think this is the case:
- Adding an
if (n == 0) n = 1
statement at the start of the loop results in both versions being as efficient, since factoring out the multiplication no longer guarantees that the result will be the same - The optimized version (by factoring out the multiplication by 2) is exactly as fast as the
2 * (i * i)
version
Here is the test code that I used to draw these conclusions:
public static void main(String args) {
long fastVersion = 0;
long slowVersion = 0;
long optimizedVersion = 0;
long modifiedFastVersion = 0;
long modifiedSlowVersion = 0;
for (int i = 0; i < 10; i++) {
fastVersion += fastVersion();
slowVersion += slowVersion();
optimizedVersion += optimizedVersion();
modifiedFastVersion += modifiedFastVersion();
modifiedSlowVersion += modifiedSlowVersion();
}
System.out.println("Fast version: " + (double) fastVersion / 1000000000 + " s");
System.out.println("Slow version: " + (double) slowVersion / 1000000000 + " s");
System.out.println("Optimized version: " + (double) optimizedVersion / 1000000000 + " s");
System.out.println("Modified fast version: " + (double) modifiedFastVersion / 1000000000 + " s");
System.out.println("Modified slow version: " + (double) modifiedSlowVersion / 1000000000 + " s");
}
private static long fastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long slowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
private static long optimizedVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
return System.nanoTime() - startTime;
}
private static long modifiedFastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long modifiedSlowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
And here are the results:
Fast version: 5.7274411 s
Slow version: 7.6190804 s
Optimized version: 5.1348007 s
Modified fast version: 7.1492705 s
Modified slow version: 7.2952668 s
When the multiplication is 2 * (i * i)
, the JVM is able to factor out the multiplication by 2
from the loop, resulting in this equivalent but more efficient code:
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
but when the multiplication is (2 * i) * i
, the JVM doesn't optimize it since the multiplication by a constant is no longer right before the addition.
Here are a few reasons why I think this is the case:
- Adding an
if (n == 0) n = 1
statement at the start of the loop results in both versions being as efficient, since factoring out the multiplication no longer guarantees that the result will be the same - The optimized version (by factoring out the multiplication by 2) is exactly as fast as the
2 * (i * i)
version
Here is the test code that I used to draw these conclusions:
public static void main(String args) {
long fastVersion = 0;
long slowVersion = 0;
long optimizedVersion = 0;
long modifiedFastVersion = 0;
long modifiedSlowVersion = 0;
for (int i = 0; i < 10; i++) {
fastVersion += fastVersion();
slowVersion += slowVersion();
optimizedVersion += optimizedVersion();
modifiedFastVersion += modifiedFastVersion();
modifiedSlowVersion += modifiedSlowVersion();
}
System.out.println("Fast version: " + (double) fastVersion / 1000000000 + " s");
System.out.println("Slow version: " + (double) slowVersion / 1000000000 + " s");
System.out.println("Optimized version: " + (double) optimizedVersion / 1000000000 + " s");
System.out.println("Modified fast version: " + (double) modifiedFastVersion / 1000000000 + " s");
System.out.println("Modified slow version: " + (double) modifiedSlowVersion / 1000000000 + " s");
}
private static long fastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long slowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
private static long optimizedVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += i * i;
}
n *= 2;
return System.nanoTime() - startTime;
}
private static long modifiedFastVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * (i * i);
}
return System.nanoTime() - startTime;
}
private static long modifiedSlowVersion() {
long startTime = System.nanoTime();
int n = 0;
for (int i = 0; i < 1000000000; i++) {
if (n == 0) n = 1;
n += 2 * i * i;
}
return System.nanoTime() - startTime;
}
And here are the results:
Fast version: 5.7274411 s
Slow version: 7.6190804 s
Optimized version: 5.1348007 s
Modified fast version: 7.1492705 s
Modified slow version: 7.2952668 s
answered Nov 23 at 21:44
Runemoro
2,58521239
2,58521239
3
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
2
I think on the optimizedVersion, it should ben *= 2000000000;
– StefansArya
Nov 24 at 1:19
4
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.
– Martin Bonner
Nov 26 at 15:57
5
The math equation was just like this2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.
– StefansArya
Nov 26 at 17:22
2
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
|
show 2 more comments
3
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
2
I think on the optimizedVersion, it should ben *= 2000000000;
– StefansArya
Nov 24 at 1:19
4
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.
– Martin Bonner
Nov 26 at 15:57
5
The math equation was just like this2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.
– StefansArya
Nov 26 at 17:22
2
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
3
3
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
here is a benchmark: github.com/jawb-software/stackoverflow-53452713
– dit
Nov 23 at 22:27
2
2
I think on the optimizedVersion, it should be
n *= 2000000000;
– StefansArya
Nov 24 at 1:19
I think on the optimizedVersion, it should be
n *= 2000000000;
– StefansArya
Nov 24 at 1:19
4
4
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate
2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating 1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.– Martin Bonner
Nov 26 at 15:57
@StefansArya - No. Consider the case where the limit is 4, and we are trying to calculate
2*1*1 + 2*2*2 + 2*3*3
. It is obvious that calculating 1*1 + 2*2 + 3*3
and multiplying by 2 is correct, whereas multiply by 8 would not be.– Martin Bonner
Nov 26 at 15:57
5
5
The math equation was just like this
2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.– StefansArya
Nov 26 at 17:22
The math equation was just like this
2(1²) + 2(2²) + 2(3²) = 2(1² + 2² + 3²)
. That was very simple and I just forgot it because the loop increment.– StefansArya
Nov 26 at 17:22
2
2
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
If you print out the assembly using a debug jvm, this does not appear to be correct. You will see a bunch of sall ... ,#1, which are multiplies by 2, in the loop. Interestingly, the slower version does not appear to have multiplies in the loop.
– Daniel Berlin
Dec 1 at 4:53
|
show 2 more comments
ByteCodes: https://cs.nyu.edu/courses/fall00/V22.0201-001/jvm2.html
ByteCodes Viewer: https://github.com/Konloch/bytecode-viewer
On my JDK (Win10 64 1.8.0_65-b17) I can reproduce and explain:
public static void main(String args) {
int repeat = 10;
long A = 0;
long B = 0;
for (int i = 0; i < repeat; i++) {
A += test();
B += testB();
}
System.out.println(A / repeat + " ms");
System.out.println(B / repeat + " ms");
}
private static long test() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multi(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multi(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms A " + n);
return ms;
}
private static long testB() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multiB(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multiB(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms B " + n);
return ms;
}
private static int multiB(int i) {
return 2 * (i * i);
}
private static int multi(int i) {
return 2 * i * i;
}
Output:
...
405 ms A 785527736
327 ms B 785527736
404 ms A 785527736
329 ms B 785527736
404 ms A 785527736
328 ms B 785527736
404 ms A 785527736
328 ms B 785527736
410 ms
333 ms
So why?
The Byte code is this:
private static multiB(int arg0) { // 2 * (i * i)
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
iload0
imul
imul
ireturn
}
L2 {
}
}
private static multi(int arg0) { // 2 * i * i
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
imul
iload0
imul
ireturn
}
L2 {
}
}
The difference being:
With brackets (2 * (i * i)
):
- push const stack
- push local on stack
- push local on stack
- multiply top of stack
- multiply top of stack
Without brackets (2 * i * i
):
- push const stack
- push local on stack
- multiply top of stack
- push local on stack
- multiply top of stack
Loading all on stack and then working back down is faster than switching between putting on stack and operating on it.
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
add a comment |
ByteCodes: https://cs.nyu.edu/courses/fall00/V22.0201-001/jvm2.html
ByteCodes Viewer: https://github.com/Konloch/bytecode-viewer
On my JDK (Win10 64 1.8.0_65-b17) I can reproduce and explain:
public static void main(String args) {
int repeat = 10;
long A = 0;
long B = 0;
for (int i = 0; i < repeat; i++) {
A += test();
B += testB();
}
System.out.println(A / repeat + " ms");
System.out.println(B / repeat + " ms");
}
private static long test() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multi(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multi(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms A " + n);
return ms;
}
private static long testB() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multiB(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multiB(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms B " + n);
return ms;
}
private static int multiB(int i) {
return 2 * (i * i);
}
private static int multi(int i) {
return 2 * i * i;
}
Output:
...
405 ms A 785527736
327 ms B 785527736
404 ms A 785527736
329 ms B 785527736
404 ms A 785527736
328 ms B 785527736
404 ms A 785527736
328 ms B 785527736
410 ms
333 ms
So why?
The Byte code is this:
private static multiB(int arg0) { // 2 * (i * i)
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
iload0
imul
imul
ireturn
}
L2 {
}
}
private static multi(int arg0) { // 2 * i * i
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
imul
iload0
imul
ireturn
}
L2 {
}
}
The difference being:
With brackets (2 * (i * i)
):
- push const stack
- push local on stack
- push local on stack
- multiply top of stack
- multiply top of stack
Without brackets (2 * i * i
):
- push const stack
- push local on stack
- multiply top of stack
- push local on stack
- multiply top of stack
Loading all on stack and then working back down is faster than switching between putting on stack and operating on it.
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
add a comment |
ByteCodes: https://cs.nyu.edu/courses/fall00/V22.0201-001/jvm2.html
ByteCodes Viewer: https://github.com/Konloch/bytecode-viewer
On my JDK (Win10 64 1.8.0_65-b17) I can reproduce and explain:
public static void main(String args) {
int repeat = 10;
long A = 0;
long B = 0;
for (int i = 0; i < repeat; i++) {
A += test();
B += testB();
}
System.out.println(A / repeat + " ms");
System.out.println(B / repeat + " ms");
}
private static long test() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multi(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multi(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms A " + n);
return ms;
}
private static long testB() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multiB(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multiB(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms B " + n);
return ms;
}
private static int multiB(int i) {
return 2 * (i * i);
}
private static int multi(int i) {
return 2 * i * i;
}
Output:
...
405 ms A 785527736
327 ms B 785527736
404 ms A 785527736
329 ms B 785527736
404 ms A 785527736
328 ms B 785527736
404 ms A 785527736
328 ms B 785527736
410 ms
333 ms
So why?
The Byte code is this:
private static multiB(int arg0) { // 2 * (i * i)
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
iload0
imul
imul
ireturn
}
L2 {
}
}
private static multi(int arg0) { // 2 * i * i
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
imul
iload0
imul
ireturn
}
L2 {
}
}
The difference being:
With brackets (2 * (i * i)
):
- push const stack
- push local on stack
- push local on stack
- multiply top of stack
- multiply top of stack
Without brackets (2 * i * i
):
- push const stack
- push local on stack
- multiply top of stack
- push local on stack
- multiply top of stack
Loading all on stack and then working back down is faster than switching between putting on stack and operating on it.
ByteCodes: https://cs.nyu.edu/courses/fall00/V22.0201-001/jvm2.html
ByteCodes Viewer: https://github.com/Konloch/bytecode-viewer
On my JDK (Win10 64 1.8.0_65-b17) I can reproduce and explain:
public static void main(String args) {
int repeat = 10;
long A = 0;
long B = 0;
for (int i = 0; i < repeat; i++) {
A += test();
B += testB();
}
System.out.println(A / repeat + " ms");
System.out.println(B / repeat + " ms");
}
private static long test() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multi(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multi(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms A " + n);
return ms;
}
private static long testB() {
int n = 0;
for (int i = 0; i < 1000; i++) {
n += multiB(i);
}
long startTime = System.currentTimeMillis();
for (int i = 0; i < 1000000000; i++) {
n += multiB(i);
}
long ms = (System.currentTimeMillis() - startTime);
System.out.println(ms + " ms B " + n);
return ms;
}
private static int multiB(int i) {
return 2 * (i * i);
}
private static int multi(int i) {
return 2 * i * i;
}
Output:
...
405 ms A 785527736
327 ms B 785527736
404 ms A 785527736
329 ms B 785527736
404 ms A 785527736
328 ms B 785527736
404 ms A 785527736
328 ms B 785527736
410 ms
333 ms
So why?
The Byte code is this:
private static multiB(int arg0) { // 2 * (i * i)
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
iload0
imul
imul
ireturn
}
L2 {
}
}
private static multi(int arg0) { // 2 * i * i
<localVar:index=0 , name=i , desc=I, sig=null, start=L1, end=L2>
L1 {
iconst_2
iload0
imul
iload0
imul
ireturn
}
L2 {
}
}
The difference being:
With brackets (2 * (i * i)
):
- push const stack
- push local on stack
- push local on stack
- multiply top of stack
- multiply top of stack
Without brackets (2 * i * i
):
- push const stack
- push local on stack
- multiply top of stack
- push local on stack
- multiply top of stack
Loading all on stack and then working back down is faster than switching between putting on stack and operating on it.
edited Nov 23 at 21:30
answered Nov 23 at 21:19
DSchmidt
587410
587410
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
add a comment |
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
But why is push-push-multiply-multiply faster than push-multiply-push-multiply?
– m0skit0
Dec 1 at 12:04
add a comment |
Kasperd asked in a comment of the accepted answer:
The Java and C examples use quite different register names. Are both example using the AMD64 ISA?
xor edx, edx
xor eax, eax
.L2:
mov ecx, edx
imul ecx, edx
add edx, 1
lea eax, [rax+rcx*2]
cmp edx, 1000000000
jne .L2
I don't have enough reputation to answer this in the comments, but these are the same ISA. It's worth pointing out that the GCC version uses 32-bit integer logic and the JVM compiled version uses 64-bit integer logic internally.
R8 to R15 are just new X86_64 registers. EAX to EDX are the lower parts of the RAX to RDX general purpose registers. The important part in the answer is that the GCC version is not unrolled. It simply executes one round of the loop per actual machine code loop. While the JVM version has 16 rounds of the loop in one physical loop (based on rustyx answer, I did not reinterpret the assembly). This is one of the reasons why there are more registers being used since the loop body is actually 16 times longer.
add a comment |
Kasperd asked in a comment of the accepted answer:
The Java and C examples use quite different register names. Are both example using the AMD64 ISA?
xor edx, edx
xor eax, eax
.L2:
mov ecx, edx
imul ecx, edx
add edx, 1
lea eax, [rax+rcx*2]
cmp edx, 1000000000
jne .L2
I don't have enough reputation to answer this in the comments, but these are the same ISA. It's worth pointing out that the GCC version uses 32-bit integer logic and the JVM compiled version uses 64-bit integer logic internally.
R8 to R15 are just new X86_64 registers. EAX to EDX are the lower parts of the RAX to RDX general purpose registers. The important part in the answer is that the GCC version is not unrolled. It simply executes one round of the loop per actual machine code loop. While the JVM version has 16 rounds of the loop in one physical loop (based on rustyx answer, I did not reinterpret the assembly). This is one of the reasons why there are more registers being used since the loop body is actually 16 times longer.
add a comment |
Kasperd asked in a comment of the accepted answer:
The Java and C examples use quite different register names. Are both example using the AMD64 ISA?
xor edx, edx
xor eax, eax
.L2:
mov ecx, edx
imul ecx, edx
add edx, 1
lea eax, [rax+rcx*2]
cmp edx, 1000000000
jne .L2
I don't have enough reputation to answer this in the comments, but these are the same ISA. It's worth pointing out that the GCC version uses 32-bit integer logic and the JVM compiled version uses 64-bit integer logic internally.
R8 to R15 are just new X86_64 registers. EAX to EDX are the lower parts of the RAX to RDX general purpose registers. The important part in the answer is that the GCC version is not unrolled. It simply executes one round of the loop per actual machine code loop. While the JVM version has 16 rounds of the loop in one physical loop (based on rustyx answer, I did not reinterpret the assembly). This is one of the reasons why there are more registers being used since the loop body is actually 16 times longer.
Kasperd asked in a comment of the accepted answer:
The Java and C examples use quite different register names. Are both example using the AMD64 ISA?
xor edx, edx
xor eax, eax
.L2:
mov ecx, edx
imul ecx, edx
add edx, 1
lea eax, [rax+rcx*2]
cmp edx, 1000000000
jne .L2
I don't have enough reputation to answer this in the comments, but these are the same ISA. It's worth pointing out that the GCC version uses 32-bit integer logic and the JVM compiled version uses 64-bit integer logic internally.
R8 to R15 are just new X86_64 registers. EAX to EDX are the lower parts of the RAX to RDX general purpose registers. The important part in the answer is that the GCC version is not unrolled. It simply executes one round of the loop per actual machine code loop. While the JVM version has 16 rounds of the loop in one physical loop (based on rustyx answer, I did not reinterpret the assembly). This is one of the reasons why there are more registers being used since the loop body is actually 16 times longer.
answered Nov 25 at 18:18
Puzzled
45632
45632
add a comment |
add a comment |
While not directly related to the question's environment, just for the curiosity, I did the same test on .Net Core 2.1, x64, release mode.
Here is the interesting result, confirming similar phonemenia (other way around) happening over the dark side of the force. Code:
static void Main(string args)
{
Stopwatch watch = new Stopwatch();
Console.WriteLine("2 * (i * i)");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * (i * i);
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
Console.WriteLine();
Console.WriteLine("2 * i * i");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * i * i;
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
}
Result:
2 * (i * i)
- result:119860736, 438ms
- result:119860736, 433ms
- result:119860736, 437ms
- result:119860736, 435ms
- result:119860736, 436ms
- result:119860736, 435ms
- result:119860736, 435ms
- result:119860736, 439ms
- result:119860736, 436ms
- result:119860736, 437ms
2 * i * i
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 418ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 416ms
- result:119860736, 417ms
- result:119860736, 418ms
1
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
1
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
1
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
5
Except this is the other way around
– leppie
Nov 30 at 14:55
2
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
|
show 4 more comments
While not directly related to the question's environment, just for the curiosity, I did the same test on .Net Core 2.1, x64, release mode.
Here is the interesting result, confirming similar phonemenia (other way around) happening over the dark side of the force. Code:
static void Main(string args)
{
Stopwatch watch = new Stopwatch();
Console.WriteLine("2 * (i * i)");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * (i * i);
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
Console.WriteLine();
Console.WriteLine("2 * i * i");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * i * i;
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
}
Result:
2 * (i * i)
- result:119860736, 438ms
- result:119860736, 433ms
- result:119860736, 437ms
- result:119860736, 435ms
- result:119860736, 436ms
- result:119860736, 435ms
- result:119860736, 435ms
- result:119860736, 439ms
- result:119860736, 436ms
- result:119860736, 437ms
2 * i * i
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 418ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 416ms
- result:119860736, 417ms
- result:119860736, 418ms
1
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
1
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
1
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
5
Except this is the other way around
– leppie
Nov 30 at 14:55
2
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
|
show 4 more comments
While not directly related to the question's environment, just for the curiosity, I did the same test on .Net Core 2.1, x64, release mode.
Here is the interesting result, confirming similar phonemenia (other way around) happening over the dark side of the force. Code:
static void Main(string args)
{
Stopwatch watch = new Stopwatch();
Console.WriteLine("2 * (i * i)");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * (i * i);
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
Console.WriteLine();
Console.WriteLine("2 * i * i");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * i * i;
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
}
Result:
2 * (i * i)
- result:119860736, 438ms
- result:119860736, 433ms
- result:119860736, 437ms
- result:119860736, 435ms
- result:119860736, 436ms
- result:119860736, 435ms
- result:119860736, 435ms
- result:119860736, 439ms
- result:119860736, 436ms
- result:119860736, 437ms
2 * i * i
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 418ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 416ms
- result:119860736, 417ms
- result:119860736, 418ms
While not directly related to the question's environment, just for the curiosity, I did the same test on .Net Core 2.1, x64, release mode.
Here is the interesting result, confirming similar phonemenia (other way around) happening over the dark side of the force. Code:
static void Main(string args)
{
Stopwatch watch = new Stopwatch();
Console.WriteLine("2 * (i * i)");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * (i * i);
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
Console.WriteLine();
Console.WriteLine("2 * i * i");
for (int a = 0; a < 10; a++)
{
int n = 0;
watch.Restart();
for (int i = 0; i < 1000000000; i++)
{
n += 2 * i * i;
}
watch.Stop();
Console.WriteLine($"result:{n}, {watch.ElapsedMilliseconds}ms");
}
}
Result:
2 * (i * i)
- result:119860736, 438ms
- result:119860736, 433ms
- result:119860736, 437ms
- result:119860736, 435ms
- result:119860736, 436ms
- result:119860736, 435ms
- result:119860736, 435ms
- result:119860736, 439ms
- result:119860736, 436ms
- result:119860736, 437ms
2 * i * i
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 418ms
- result:119860736, 417ms
- result:119860736, 418ms
- result:119860736, 416ms
- result:119860736, 417ms
- result:119860736, 418ms
edited Dec 3 at 11:11
answered Nov 28 at 8:12
Ünsal Ersöz
32224
32224
1
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
1
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
1
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
5
Except this is the other way around
– leppie
Nov 30 at 14:55
2
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
|
show 4 more comments
1
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
1
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
1
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
5
Except this is the other way around
– leppie
Nov 30 at 14:55
2
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
1
1
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
While this isn't an answer to the question, it does add value. That being said, if something is vital to your post, please in-line it in the post rather than linking to an off-site resource. Links go dead.
– Jared Smith
Nov 28 at 13:54
1
1
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
@JaredSmith Thanks for the feedback. Considering the link you mention is the "result" link, that image is not an off-site source. I uploaded it to the stackoverflow via its own panel.
– Ünsal Ersöz
Nov 28 at 14:32
1
1
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
It's a link to imgur, so yes, it is, it doesn't matter how you added the link. I fail to see what's so difficult about copy-pasting some console output.
– Jared Smith
Nov 28 at 14:49
5
5
Except this is the other way around
– leppie
Nov 30 at 14:55
Except this is the other way around
– leppie
Nov 30 at 14:55
2
2
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
@SamB it's still on the imgur.com domain, which means it'll survive only for as long as imgur.
– p91paul
Dec 1 at 10:22
|
show 4 more comments
I got similar results:
2 * (i * i): 0.458765943 s, n=119860736
2 * i * i: 0.580255126 s, n=119860736
I got the SAME results if both loops were in the same program, or each was in a separate .java file/.class, executed on a separate run.
Finally, here is a javap -c -v <.java>
decompile of each:
3: ldc #3 // String 2 * (i * i):
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: iload 4
30: imul
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
vs.
3: ldc #3 // String 2 * i * i:
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: imul
29: iload 4
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
FYI -
java -version
java version "1.8.0_121"
Java(TM) SE Runtime Environment (build 1.8.0_121-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.121-b13, mixed mode)
1
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
2
Get a debug jre and run with-XX:+PrintOptoAssembly
. Or just use vtune or alike.
– rustyx
Nov 23 at 22:42
1
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
|
show 2 more comments
I got similar results:
2 * (i * i): 0.458765943 s, n=119860736
2 * i * i: 0.580255126 s, n=119860736
I got the SAME results if both loops were in the same program, or each was in a separate .java file/.class, executed on a separate run.
Finally, here is a javap -c -v <.java>
decompile of each:
3: ldc #3 // String 2 * (i * i):
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: iload 4
30: imul
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
vs.
3: ldc #3 // String 2 * i * i:
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: imul
29: iload 4
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
FYI -
java -version
java version "1.8.0_121"
Java(TM) SE Runtime Environment (build 1.8.0_121-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.121-b13, mixed mode)
1
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
2
Get a debug jre and run with-XX:+PrintOptoAssembly
. Or just use vtune or alike.
– rustyx
Nov 23 at 22:42
1
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
|
show 2 more comments
I got similar results:
2 * (i * i): 0.458765943 s, n=119860736
2 * i * i: 0.580255126 s, n=119860736
I got the SAME results if both loops were in the same program, or each was in a separate .java file/.class, executed on a separate run.
Finally, here is a javap -c -v <.java>
decompile of each:
3: ldc #3 // String 2 * (i * i):
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: iload 4
30: imul
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
vs.
3: ldc #3 // String 2 * i * i:
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: imul
29: iload 4
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
FYI -
java -version
java version "1.8.0_121"
Java(TM) SE Runtime Environment (build 1.8.0_121-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.121-b13, mixed mode)
I got similar results:
2 * (i * i): 0.458765943 s, n=119860736
2 * i * i: 0.580255126 s, n=119860736
I got the SAME results if both loops were in the same program, or each was in a separate .java file/.class, executed on a separate run.
Finally, here is a javap -c -v <.java>
decompile of each:
3: ldc #3 // String 2 * (i * i):
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: iload 4
30: imul
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
vs.
3: ldc #3 // String 2 * i * i:
5: invokevirtual #4 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
8: invokestatic #5 // Method java/lang/System.nanoTime:()J
11: lstore_1
12: iconst_0
13: istore_3
14: iconst_0
15: istore 4
17: iload 4
19: ldc #6 // int 1000000000
21: if_icmpge 40
24: iload_3
25: iconst_2
26: iload 4
28: imul
29: iload 4
31: imul
32: iadd
33: istore_3
34: iinc 4, 1
37: goto 17
FYI -
java -version
java version "1.8.0_121"
Java(TM) SE Runtime Environment (build 1.8.0_121-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.121-b13, mixed mode)
answered Nov 23 at 21:10
paulsm4
77k999126
77k999126
1
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
2
Get a debug jre and run with-XX:+PrintOptoAssembly
. Or just use vtune or alike.
– rustyx
Nov 23 at 22:42
1
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
|
show 2 more comments
1
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
2
Get a debug jre and run with-XX:+PrintOptoAssembly
. Or just use vtune or alike.
– rustyx
Nov 23 at 22:42
1
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
1
1
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
A better answer and maybe you can vote to undelete - stackoverflow.com/a/53452836/1746118 ... Side note - I am not the downvoter anyway.
– nullpointer
Nov 23 at 21:11
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
@nullpointer - I agree. I'd definitely vote to undelete, if I could. I'd also like to "double upvote" stefan for giving a quantitative definition of "significant"
– paulsm4
Nov 23 at 21:14
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
That one was self-deleted since it measured the wrong thing - see that author's comment on the question above
– Krease
Nov 23 at 21:16
2
2
Get a debug jre and run with
-XX:+PrintOptoAssembly
. Or just use vtune or alike.– rustyx
Nov 23 at 22:42
Get a debug jre and run with
-XX:+PrintOptoAssembly
. Or just use vtune or alike.– rustyx
Nov 23 at 22:42
1
1
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
@ rustyx - If the problem is the JIT implementation ... then "getting a debug version" OF A COMPLETELY DIFFERENT JRE isn't necessarily going to help. Nevertheless: it sounds like what you found above with your JIT disassembly on your JRE also explains the behavior on the OP's JRE and mine. And also explains why other JRE's behave "differently". +1: thank you for the excellent detective work!
– paulsm4
Nov 24 at 8:06
|
show 2 more comments
I tried a JMH using the default archetype: I also added optimized version based Runemoro' explanation .
@State(Scope.Benchmark)
@Warmup(iterations = 2)
@Fork(1)
@Measurement(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
//@BenchmarkMode({ Mode.All })
@BenchmarkMode(Mode.AverageTime)
public class MyBenchmark {
@Param({ "100", "1000", "1000000000" })
private int size;
@Benchmark
public int two_square_i() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
@Benchmark
public int square_i_two() {
int n = 0;
for (int i = 0; i < size; i++) {
n += i * i;
}
return 2*n;
}
@Benchmark
public int two_i_() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
}
The result are here:
Benchmark (size) Mode Samples Score Score error Units
o.s.MyBenchmark.square_i_two 100 avgt 10 58,062 1,410 ns/op
o.s.MyBenchmark.square_i_two 1000 avgt 10 547,393 12,851 ns/op
o.s.MyBenchmark.square_i_two 1000000000 avgt 10 540343681,267 16795210,324 ns/op
o.s.MyBenchmark.two_i_ 100 avgt 10 87,491 2,004 ns/op
o.s.MyBenchmark.two_i_ 1000 avgt 10 1015,388 30,313 ns/op
o.s.MyBenchmark.two_i_ 1000000000 avgt 10 967100076,600 24929570,556 ns/op
o.s.MyBenchmark.two_square_i 100 avgt 10 70,715 2,107 ns/op
o.s.MyBenchmark.two_square_i 1000 avgt 10 686,977 24,613 ns/op
o.s.MyBenchmark.two_square_i 1000000000 avgt 10 652736811,450 27015580,488 ns/op
On my PC (Core i7 860, doing nothing much apart reading on my smartphone):
n += i*i
thenn*2
is first
2 * (i * i)
is second.
The JVM is clearly not optimizing the same way than a human does (based on Runemoro answer).
Now then, reading bytecode: javap -c -v ./target/classes/org/sample/MyBenchmark.class
- Differences between 2*(i*i) (left) and 2*i*i (right) here: https://www.diffchecker.com/cvSFppWI
- Differences between 2*(i*i) and the optimized version here: https://www.diffchecker.com/I1XFu5dP
I am not expert on bytecode but we iload_2
before we imul
: that's probably where you get the difference: I can suppose that the JVM optimize reading i
twice (i
is already here, there is no need to load it again) whilst in the 2*i*i
it can't.
3
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
add a comment |
I tried a JMH using the default archetype: I also added optimized version based Runemoro' explanation .
@State(Scope.Benchmark)
@Warmup(iterations = 2)
@Fork(1)
@Measurement(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
//@BenchmarkMode({ Mode.All })
@BenchmarkMode(Mode.AverageTime)
public class MyBenchmark {
@Param({ "100", "1000", "1000000000" })
private int size;
@Benchmark
public int two_square_i() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
@Benchmark
public int square_i_two() {
int n = 0;
for (int i = 0; i < size; i++) {
n += i * i;
}
return 2*n;
}
@Benchmark
public int two_i_() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
}
The result are here:
Benchmark (size) Mode Samples Score Score error Units
o.s.MyBenchmark.square_i_two 100 avgt 10 58,062 1,410 ns/op
o.s.MyBenchmark.square_i_two 1000 avgt 10 547,393 12,851 ns/op
o.s.MyBenchmark.square_i_two 1000000000 avgt 10 540343681,267 16795210,324 ns/op
o.s.MyBenchmark.two_i_ 100 avgt 10 87,491 2,004 ns/op
o.s.MyBenchmark.two_i_ 1000 avgt 10 1015,388 30,313 ns/op
o.s.MyBenchmark.two_i_ 1000000000 avgt 10 967100076,600 24929570,556 ns/op
o.s.MyBenchmark.two_square_i 100 avgt 10 70,715 2,107 ns/op
o.s.MyBenchmark.two_square_i 1000 avgt 10 686,977 24,613 ns/op
o.s.MyBenchmark.two_square_i 1000000000 avgt 10 652736811,450 27015580,488 ns/op
On my PC (Core i7 860, doing nothing much apart reading on my smartphone):
n += i*i
thenn*2
is first
2 * (i * i)
is second.
The JVM is clearly not optimizing the same way than a human does (based on Runemoro answer).
Now then, reading bytecode: javap -c -v ./target/classes/org/sample/MyBenchmark.class
- Differences between 2*(i*i) (left) and 2*i*i (right) here: https://www.diffchecker.com/cvSFppWI
- Differences between 2*(i*i) and the optimized version here: https://www.diffchecker.com/I1XFu5dP
I am not expert on bytecode but we iload_2
before we imul
: that's probably where you get the difference: I can suppose that the JVM optimize reading i
twice (i
is already here, there is no need to load it again) whilst in the 2*i*i
it can't.
3
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
add a comment |
I tried a JMH using the default archetype: I also added optimized version based Runemoro' explanation .
@State(Scope.Benchmark)
@Warmup(iterations = 2)
@Fork(1)
@Measurement(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
//@BenchmarkMode({ Mode.All })
@BenchmarkMode(Mode.AverageTime)
public class MyBenchmark {
@Param({ "100", "1000", "1000000000" })
private int size;
@Benchmark
public int two_square_i() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
@Benchmark
public int square_i_two() {
int n = 0;
for (int i = 0; i < size; i++) {
n += i * i;
}
return 2*n;
}
@Benchmark
public int two_i_() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
}
The result are here:
Benchmark (size) Mode Samples Score Score error Units
o.s.MyBenchmark.square_i_two 100 avgt 10 58,062 1,410 ns/op
o.s.MyBenchmark.square_i_two 1000 avgt 10 547,393 12,851 ns/op
o.s.MyBenchmark.square_i_two 1000000000 avgt 10 540343681,267 16795210,324 ns/op
o.s.MyBenchmark.two_i_ 100 avgt 10 87,491 2,004 ns/op
o.s.MyBenchmark.two_i_ 1000 avgt 10 1015,388 30,313 ns/op
o.s.MyBenchmark.two_i_ 1000000000 avgt 10 967100076,600 24929570,556 ns/op
o.s.MyBenchmark.two_square_i 100 avgt 10 70,715 2,107 ns/op
o.s.MyBenchmark.two_square_i 1000 avgt 10 686,977 24,613 ns/op
o.s.MyBenchmark.two_square_i 1000000000 avgt 10 652736811,450 27015580,488 ns/op
On my PC (Core i7 860, doing nothing much apart reading on my smartphone):
n += i*i
thenn*2
is first
2 * (i * i)
is second.
The JVM is clearly not optimizing the same way than a human does (based on Runemoro answer).
Now then, reading bytecode: javap -c -v ./target/classes/org/sample/MyBenchmark.class
- Differences between 2*(i*i) (left) and 2*i*i (right) here: https://www.diffchecker.com/cvSFppWI
- Differences between 2*(i*i) and the optimized version here: https://www.diffchecker.com/I1XFu5dP
I am not expert on bytecode but we iload_2
before we imul
: that's probably where you get the difference: I can suppose that the JVM optimize reading i
twice (i
is already here, there is no need to load it again) whilst in the 2*i*i
it can't.
I tried a JMH using the default archetype: I also added optimized version based Runemoro' explanation .
@State(Scope.Benchmark)
@Warmup(iterations = 2)
@Fork(1)
@Measurement(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
//@BenchmarkMode({ Mode.All })
@BenchmarkMode(Mode.AverageTime)
public class MyBenchmark {
@Param({ "100", "1000", "1000000000" })
private int size;
@Benchmark
public int two_square_i() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
@Benchmark
public int square_i_two() {
int n = 0;
for (int i = 0; i < size; i++) {
n += i * i;
}
return 2*n;
}
@Benchmark
public int two_i_() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
}
The result are here:
Benchmark (size) Mode Samples Score Score error Units
o.s.MyBenchmark.square_i_two 100 avgt 10 58,062 1,410 ns/op
o.s.MyBenchmark.square_i_two 1000 avgt 10 547,393 12,851 ns/op
o.s.MyBenchmark.square_i_two 1000000000 avgt 10 540343681,267 16795210,324 ns/op
o.s.MyBenchmark.two_i_ 100 avgt 10 87,491 2,004 ns/op
o.s.MyBenchmark.two_i_ 1000 avgt 10 1015,388 30,313 ns/op
o.s.MyBenchmark.two_i_ 1000000000 avgt 10 967100076,600 24929570,556 ns/op
o.s.MyBenchmark.two_square_i 100 avgt 10 70,715 2,107 ns/op
o.s.MyBenchmark.two_square_i 1000 avgt 10 686,977 24,613 ns/op
o.s.MyBenchmark.two_square_i 1000000000 avgt 10 652736811,450 27015580,488 ns/op
On my PC (Core i7 860, doing nothing much apart reading on my smartphone):
n += i*i
thenn*2
is first
2 * (i * i)
is second.
The JVM is clearly not optimizing the same way than a human does (based on Runemoro answer).
Now then, reading bytecode: javap -c -v ./target/classes/org/sample/MyBenchmark.class
- Differences between 2*(i*i) (left) and 2*i*i (right) here: https://www.diffchecker.com/cvSFppWI
- Differences between 2*(i*i) and the optimized version here: https://www.diffchecker.com/I1XFu5dP
I am not expert on bytecode but we iload_2
before we imul
: that's probably where you get the difference: I can suppose that the JVM optimize reading i
twice (i
is already here, there is no need to load it again) whilst in the 2*i*i
it can't.
answered Nov 23 at 22:10
NoDataFound
5,6671740
5,6671740
3
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
add a comment |
3
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
3
3
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
AFAICT bytecode is pretty irrelevant for performance, and I wouldn't try to estimate what's faster based on it. It's just the source code for the JIT compiler... sure can meaning-preserving reordering source code lines change the resulting code and it's efficiency, but that all pretty unpredictable.
– maaartinus
Nov 26 at 2:33
add a comment |
More of an addendum. I did repro the experiment using the latest Java 8 JVM from IBM:
java version "1.8.0_191"
Java(TM) 2 Runtime Environment, Standard Edition (IBM build 1.8.0_191-b12 26_Oct_2018_18_45 Mac OS X x64(SR5 FP25))
Java HotSpot(TM) 64-Bit Server VM (build 25.191-b12, mixed mode)
and this shows very similar results:
0.374653912 s
n = 119860736
0.447778698 s
n = 119860736
( second results using 2 * i * i ).
Interestingly enough, when running on the same machine, but using Oracle java:
Java version "1.8.0_181"
Java(TM) SE Runtime Environment (build 1.8.0_181-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.181-b13, mixed mode)
results are on average a bit slower:
0.414331815 s
n = 119860736
0.491430656 s
n = 119860736
Long story short: even the minor version number of HotSpot matter here, as subtle differences within the JIT implementation can have notable effects.
add a comment |
More of an addendum. I did repro the experiment using the latest Java 8 JVM from IBM:
java version "1.8.0_191"
Java(TM) 2 Runtime Environment, Standard Edition (IBM build 1.8.0_191-b12 26_Oct_2018_18_45 Mac OS X x64(SR5 FP25))
Java HotSpot(TM) 64-Bit Server VM (build 25.191-b12, mixed mode)
and this shows very similar results:
0.374653912 s
n = 119860736
0.447778698 s
n = 119860736
( second results using 2 * i * i ).
Interestingly enough, when running on the same machine, but using Oracle java:
Java version "1.8.0_181"
Java(TM) SE Runtime Environment (build 1.8.0_181-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.181-b13, mixed mode)
results are on average a bit slower:
0.414331815 s
n = 119860736
0.491430656 s
n = 119860736
Long story short: even the minor version number of HotSpot matter here, as subtle differences within the JIT implementation can have notable effects.
add a comment |
More of an addendum. I did repro the experiment using the latest Java 8 JVM from IBM:
java version "1.8.0_191"
Java(TM) 2 Runtime Environment, Standard Edition (IBM build 1.8.0_191-b12 26_Oct_2018_18_45 Mac OS X x64(SR5 FP25))
Java HotSpot(TM) 64-Bit Server VM (build 25.191-b12, mixed mode)
and this shows very similar results:
0.374653912 s
n = 119860736
0.447778698 s
n = 119860736
( second results using 2 * i * i ).
Interestingly enough, when running on the same machine, but using Oracle java:
Java version "1.8.0_181"
Java(TM) SE Runtime Environment (build 1.8.0_181-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.181-b13, mixed mode)
results are on average a bit slower:
0.414331815 s
n = 119860736
0.491430656 s
n = 119860736
Long story short: even the minor version number of HotSpot matter here, as subtle differences within the JIT implementation can have notable effects.
More of an addendum. I did repro the experiment using the latest Java 8 JVM from IBM:
java version "1.8.0_191"
Java(TM) 2 Runtime Environment, Standard Edition (IBM build 1.8.0_191-b12 26_Oct_2018_18_45 Mac OS X x64(SR5 FP25))
Java HotSpot(TM) 64-Bit Server VM (build 25.191-b12, mixed mode)
and this shows very similar results:
0.374653912 s
n = 119860736
0.447778698 s
n = 119860736
( second results using 2 * i * i ).
Interestingly enough, when running on the same machine, but using Oracle java:
Java version "1.8.0_181"
Java(TM) SE Runtime Environment (build 1.8.0_181-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.181-b13, mixed mode)
results are on average a bit slower:
0.414331815 s
n = 119860736
0.491430656 s
n = 119860736
Long story short: even the minor version number of HotSpot matter here, as subtle differences within the JIT implementation can have notable effects.
answered Nov 30 at 21:07
GhostCat
88.1k1684144
88.1k1684144
add a comment |
add a comment |
Interesting observation using Java 11 and switching off loop unrolling with the following VM option:
-XX:LoopUnrollLimit=0
The loop with the 2 * (i * i)
expression results in a more compact native code1:
L0001: add eax,r11d
inc r8d
mov r11d,r8d
imul r11d,r8d
shl r11d,1h
cmp r8d,r10d
jl L0001
in comparison with the 2 * i * i
version:
L0001: add eax,r11d
mov r11d,r8d
shl r11d,1h
add r11d,2h
inc r8d
imul r11d,r8d
cmp r8d,r10d
jl L0001
Java version:
java version "11" 2018-09-25
Java(TM) SE Runtime Environment 18.9 (build 11+28)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11+28, mixed mode)
Benchmark results:
Benchmark (size) Mode Cnt Score Error Units
LoopTest.fast 1000000000 avgt 5 694,868 ± 36,470 ms/op
LoopTest.slow 1000000000 avgt 5 769,840 ± 135,006 ms/op
Benchmark source code:
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@Warmup(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@State(Scope.Thread)
@Fork(1)
public class LoopTest {
@Param("1000000000") private int size;
public static void main(String args) throws RunnerException {
Options opt =
new OptionsBuilder().include(LoopTest.class.getSimpleName())
.jvmArgs("-XX:LoopUnrollLimit=0")
.build();
new Runner(opt).run();
}
@Benchmark
public int slow() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int fast() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
}
1 - VM options used: -XX:+UnlockDiagnosticVMOptions -XX:+PrintAssembly -XX:LoopUnrollLimit=0
2
Wow, that's some braindead asm. Instead of incrementingi
before copying it to calculate2*i
, it does it after so it needs an extraadd r11d,2
instruction. (Plus it misses theadd same,same
peephole instead ofshl
by 1 (add runs on more ports). It also misses an LEA peephole forx*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.
– Peter Cordes
Dec 2 at 2:50
2
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)
– Peter Cordes
Dec 2 at 2:51
add a comment |
Interesting observation using Java 11 and switching off loop unrolling with the following VM option:
-XX:LoopUnrollLimit=0
The loop with the 2 * (i * i)
expression results in a more compact native code1:
L0001: add eax,r11d
inc r8d
mov r11d,r8d
imul r11d,r8d
shl r11d,1h
cmp r8d,r10d
jl L0001
in comparison with the 2 * i * i
version:
L0001: add eax,r11d
mov r11d,r8d
shl r11d,1h
add r11d,2h
inc r8d
imul r11d,r8d
cmp r8d,r10d
jl L0001
Java version:
java version "11" 2018-09-25
Java(TM) SE Runtime Environment 18.9 (build 11+28)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11+28, mixed mode)
Benchmark results:
Benchmark (size) Mode Cnt Score Error Units
LoopTest.fast 1000000000 avgt 5 694,868 ± 36,470 ms/op
LoopTest.slow 1000000000 avgt 5 769,840 ± 135,006 ms/op
Benchmark source code:
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@Warmup(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@State(Scope.Thread)
@Fork(1)
public class LoopTest {
@Param("1000000000") private int size;
public static void main(String args) throws RunnerException {
Options opt =
new OptionsBuilder().include(LoopTest.class.getSimpleName())
.jvmArgs("-XX:LoopUnrollLimit=0")
.build();
new Runner(opt).run();
}
@Benchmark
public int slow() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int fast() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
}
1 - VM options used: -XX:+UnlockDiagnosticVMOptions -XX:+PrintAssembly -XX:LoopUnrollLimit=0
2
Wow, that's some braindead asm. Instead of incrementingi
before copying it to calculate2*i
, it does it after so it needs an extraadd r11d,2
instruction. (Plus it misses theadd same,same
peephole instead ofshl
by 1 (add runs on more ports). It also misses an LEA peephole forx*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.
– Peter Cordes
Dec 2 at 2:50
2
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)
– Peter Cordes
Dec 2 at 2:51
add a comment |
Interesting observation using Java 11 and switching off loop unrolling with the following VM option:
-XX:LoopUnrollLimit=0
The loop with the 2 * (i * i)
expression results in a more compact native code1:
L0001: add eax,r11d
inc r8d
mov r11d,r8d
imul r11d,r8d
shl r11d,1h
cmp r8d,r10d
jl L0001
in comparison with the 2 * i * i
version:
L0001: add eax,r11d
mov r11d,r8d
shl r11d,1h
add r11d,2h
inc r8d
imul r11d,r8d
cmp r8d,r10d
jl L0001
Java version:
java version "11" 2018-09-25
Java(TM) SE Runtime Environment 18.9 (build 11+28)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11+28, mixed mode)
Benchmark results:
Benchmark (size) Mode Cnt Score Error Units
LoopTest.fast 1000000000 avgt 5 694,868 ± 36,470 ms/op
LoopTest.slow 1000000000 avgt 5 769,840 ± 135,006 ms/op
Benchmark source code:
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@Warmup(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@State(Scope.Thread)
@Fork(1)
public class LoopTest {
@Param("1000000000") private int size;
public static void main(String args) throws RunnerException {
Options opt =
new OptionsBuilder().include(LoopTest.class.getSimpleName())
.jvmArgs("-XX:LoopUnrollLimit=0")
.build();
new Runner(opt).run();
}
@Benchmark
public int slow() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int fast() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
}
1 - VM options used: -XX:+UnlockDiagnosticVMOptions -XX:+PrintAssembly -XX:LoopUnrollLimit=0
Interesting observation using Java 11 and switching off loop unrolling with the following VM option:
-XX:LoopUnrollLimit=0
The loop with the 2 * (i * i)
expression results in a more compact native code1:
L0001: add eax,r11d
inc r8d
mov r11d,r8d
imul r11d,r8d
shl r11d,1h
cmp r8d,r10d
jl L0001
in comparison with the 2 * i * i
version:
L0001: add eax,r11d
mov r11d,r8d
shl r11d,1h
add r11d,2h
inc r8d
imul r11d,r8d
cmp r8d,r10d
jl L0001
Java version:
java version "11" 2018-09-25
Java(TM) SE Runtime Environment 18.9 (build 11+28)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11+28, mixed mode)
Benchmark results:
Benchmark (size) Mode Cnt Score Error Units
LoopTest.fast 1000000000 avgt 5 694,868 ± 36,470 ms/op
LoopTest.slow 1000000000 avgt 5 769,840 ± 135,006 ms/op
Benchmark source code:
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@Warmup(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 5, time = 5, timeUnit = TimeUnit.SECONDS)
@State(Scope.Thread)
@Fork(1)
public class LoopTest {
@Param("1000000000") private int size;
public static void main(String args) throws RunnerException {
Options opt =
new OptionsBuilder().include(LoopTest.class.getSimpleName())
.jvmArgs("-XX:LoopUnrollLimit=0")
.build();
new Runner(opt).run();
}
@Benchmark
public int slow() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int fast() {
int n = 0;
for (int i = 0; i < size; i++) {
n += 2 * (i * i);
}
return n;
}
}
1 - VM options used: -XX:+UnlockDiagnosticVMOptions -XX:+PrintAssembly -XX:LoopUnrollLimit=0
edited Dec 3 at 19:23
answered Dec 2 at 2:38
Oleksandr
8,77343769
8,77343769
2
Wow, that's some braindead asm. Instead of incrementingi
before copying it to calculate2*i
, it does it after so it needs an extraadd r11d,2
instruction. (Plus it misses theadd same,same
peephole instead ofshl
by 1 (add runs on more ports). It also misses an LEA peephole forx*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.
– Peter Cordes
Dec 2 at 2:50
2
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)
– Peter Cordes
Dec 2 at 2:51
add a comment |
2
Wow, that's some braindead asm. Instead of incrementingi
before copying it to calculate2*i
, it does it after so it needs an extraadd r11d,2
instruction. (Plus it misses theadd same,same
peephole instead ofshl
by 1 (add runs on more ports). It also misses an LEA peephole forx*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.
– Peter Cordes
Dec 2 at 2:50
2
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)
– Peter Cordes
Dec 2 at 2:51
2
2
Wow, that's some braindead asm. Instead of incrementing
i
before copying it to calculate 2*i
, it does it after so it needs an extra add r11d,2
instruction. (Plus it misses the add same,same
peephole instead of shl
by 1 (add runs on more ports). It also misses an LEA peephole for x*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.– Peter Cordes
Dec 2 at 2:50
Wow, that's some braindead asm. Instead of incrementing
i
before copying it to calculate 2*i
, it does it after so it needs an extra add r11d,2
instruction. (Plus it misses the add same,same
peephole instead of shl
by 1 (add runs on more ports). It also misses an LEA peephole for x*2 + 2
(lea r11d, [r8*2 + 2]
) if it really wants to do things in that order for some crazy instruction-scheduling reason. We could already see from the unrolled version that missing out on LEA was costing it a lot of uops, same as both loops here.– Peter Cordes
Dec 2 at 2:50
2
2
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)– Peter Cordes
Dec 2 at 2:51
lea eax, [rax + r11 * 2]
would replace 2 instructions (in both loops) if the JIT compiler had time to look for that optimization in long-running loops. Any decent ahead-of-time compiler would find it. (Unless maybe tuning only for AMD, where scaled-index LEA has 2 cycle latency so maybe not worth it.)– Peter Cordes
Dec 2 at 2:51
add a comment |
The two methods of adding do generate slightly different byte code:
17: iconst_2
18: iload 4
20: iload 4
22: imul
23: imul
24: iadd
For 2 * (i * i)
vs:
17: iconst_2
18: iload 4
20: imul
21: iload 4
23: imul
24: iadd
For 2 * i * i
.
And when using a JMH benchmark like this:
@Warmup(iterations = 5, batchSize = 1)
@Measurement(iterations = 5, batchSize = 1)
@Fork(1)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@State(Scope.Benchmark)
public class MyBenchmark {
@Benchmark
public int noBrackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int brackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return n;
}
}
The difference is clear:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: <none>
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 380.889 ± 58.011 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 512.464 ± 11.098 ms/op
What you observe is correct, and not just an anomaly of your benchmarking style (i.e. no warmup, see How do I write a correct micro-benchmark in Java?)
Running again with Graal:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: -XX:+UnlockExperimentalVMOptions -XX:+EnableJVMCI -XX:+UseJVMCICompiler
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 335.100 ± 23.085 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 331.163 ± 50.670 ms/op
You see that the results are much closer, which makes sense, since Graal is an overall better performing, more modern, compiler.
So this is really just up to how well the JIT compiler is able to optimize a particular piece of code, and doesn't necessarily have a logical reason to it.
uhrrr.. yournoBrackets
is using brackets..
– Krease
Nov 23 at 20:55
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
add a comment |
The two methods of adding do generate slightly different byte code:
17: iconst_2
18: iload 4
20: iload 4
22: imul
23: imul
24: iadd
For 2 * (i * i)
vs:
17: iconst_2
18: iload 4
20: imul
21: iload 4
23: imul
24: iadd
For 2 * i * i
.
And when using a JMH benchmark like this:
@Warmup(iterations = 5, batchSize = 1)
@Measurement(iterations = 5, batchSize = 1)
@Fork(1)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@State(Scope.Benchmark)
public class MyBenchmark {
@Benchmark
public int noBrackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int brackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return n;
}
}
The difference is clear:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: <none>
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 380.889 ± 58.011 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 512.464 ± 11.098 ms/op
What you observe is correct, and not just an anomaly of your benchmarking style (i.e. no warmup, see How do I write a correct micro-benchmark in Java?)
Running again with Graal:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: -XX:+UnlockExperimentalVMOptions -XX:+EnableJVMCI -XX:+UseJVMCICompiler
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 335.100 ± 23.085 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 331.163 ± 50.670 ms/op
You see that the results are much closer, which makes sense, since Graal is an overall better performing, more modern, compiler.
So this is really just up to how well the JIT compiler is able to optimize a particular piece of code, and doesn't necessarily have a logical reason to it.
uhrrr.. yournoBrackets
is using brackets..
– Krease
Nov 23 at 20:55
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
add a comment |
The two methods of adding do generate slightly different byte code:
17: iconst_2
18: iload 4
20: iload 4
22: imul
23: imul
24: iadd
For 2 * (i * i)
vs:
17: iconst_2
18: iload 4
20: imul
21: iload 4
23: imul
24: iadd
For 2 * i * i
.
And when using a JMH benchmark like this:
@Warmup(iterations = 5, batchSize = 1)
@Measurement(iterations = 5, batchSize = 1)
@Fork(1)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@State(Scope.Benchmark)
public class MyBenchmark {
@Benchmark
public int noBrackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int brackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return n;
}
}
The difference is clear:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: <none>
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 380.889 ± 58.011 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 512.464 ± 11.098 ms/op
What you observe is correct, and not just an anomaly of your benchmarking style (i.e. no warmup, see How do I write a correct micro-benchmark in Java?)
Running again with Graal:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: -XX:+UnlockExperimentalVMOptions -XX:+EnableJVMCI -XX:+UseJVMCICompiler
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 335.100 ± 23.085 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 331.163 ± 50.670 ms/op
You see that the results are much closer, which makes sense, since Graal is an overall better performing, more modern, compiler.
So this is really just up to how well the JIT compiler is able to optimize a particular piece of code, and doesn't necessarily have a logical reason to it.
The two methods of adding do generate slightly different byte code:
17: iconst_2
18: iload 4
20: iload 4
22: imul
23: imul
24: iadd
For 2 * (i * i)
vs:
17: iconst_2
18: iload 4
20: imul
21: iload 4
23: imul
24: iadd
For 2 * i * i
.
And when using a JMH benchmark like this:
@Warmup(iterations = 5, batchSize = 1)
@Measurement(iterations = 5, batchSize = 1)
@Fork(1)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@State(Scope.Benchmark)
public class MyBenchmark {
@Benchmark
public int noBrackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * i * i;
}
return n;
}
@Benchmark
public int brackets() {
int n = 0;
for (int i = 0; i < 1000000000; i++) {
n += 2 * (i * i);
}
return n;
}
}
The difference is clear:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: <none>
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 380.889 ± 58.011 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 512.464 ± 11.098 ms/op
What you observe is correct, and not just an anomaly of your benchmarking style (i.e. no warmup, see How do I write a correct micro-benchmark in Java?)
Running again with Graal:
# JMH version: 1.21
# VM version: JDK 11, Java HotSpot(TM) 64-Bit Server VM, 11+28
# VM options: -XX:+UnlockExperimentalVMOptions -XX:+EnableJVMCI -XX:+UseJVMCICompiler
Benchmark (n) Mode Cnt Score Error Units
MyBenchmark.brackets 1000000000 avgt 5 335.100 ± 23.085 ms/op
MyBenchmark.noBrackets 1000000000 avgt 5 331.163 ± 50.670 ms/op
You see that the results are much closer, which makes sense, since Graal is an overall better performing, more modern, compiler.
So this is really just up to how well the JIT compiler is able to optimize a particular piece of code, and doesn't necessarily have a logical reason to it.
edited Dec 23 at 14:04
answered Nov 23 at 20:54
Jorn Vernee
19.8k33557
19.8k33557
uhrrr.. yournoBrackets
is using brackets..
– Krease
Nov 23 at 20:55
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
add a comment |
uhrrr.. yournoBrackets
is using brackets..
– Krease
Nov 23 at 20:55
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
uhrrr.. your
noBrackets
is using brackets..– Krease
Nov 23 at 20:55
uhrrr.. your
noBrackets
is using brackets..– Krease
Nov 23 at 20:55
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
@Krease Oops, let me fix that.
– Jorn Vernee
Nov 23 at 20:56
add a comment |
protected by Cassio Mazzochi Molin Dec 7 at 10:23
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4
I get similar results (slightly different numbers, but definitely noticeable and consistent gap, definitely more than sampling error)
– Krease
Nov 23 at 20:47
27
Also please see: stackoverflow.com/questions/504103/…
– lexicore
Nov 23 at 20:56
2
@Krease Good that you caught my mistake. According to the new benchmark I ran
2 * i * i
is slower. I'll try running with Graal as well.– Jorn Vernee
Nov 23 at 21:07
5
@nullpointer To find out for real why one is faster than the other, we'd have to get the disassembly or Ideal graphs for those methods. The assembler is very annoying to try and figure out, so I'm trying to get an OpenJDK debug build which can output nice graphs.
– Jorn Vernee
Nov 23 at 21:29
4
You could rename your question to "Why is
i * i * 2
faster than2 * i * i
?" for improved clarity that the issue is on the order of the operations.– Cœur
Nov 28 at 4:02