Adding two IEEE754 floating-point representations and interpreting the result.
This isn't for any class or homework. As part of my personal study, I'm trying to better understand the IEEE754 representation of decimal floating-point numbers in binary. I'd like to add two numbers: $1.111$ and $2.222$, then compare the result by converting the IEEE754 representation of the sum back to decimal.
Per this online tool:
- $1.111 = 00111111100011100011010100111111$
- $2.222 = 01000000000011100011010100111111$
Summing these two together using signed binary addition, I get:
$0111 1111 1001 1100 0110 1010 0111 1110$
In hexadecimal, this is:
$7F9C6A7E$
And according to this other version of the tool, that corresponds to $NaN$.
What's going on here?
binary floating-point
asked Nov 25 at 0:53
AleksandrH
1,22221123
|
show 4 more comments
This isn't for any class or homework. As part of my personal study, I'm trying to better understand the IEEE754 representation of decimal floating-point numbers in binary. I'd like to add two numbers: $1.111$ and $2.222$, then compare the result by converting the IEEE754 representation of the sum back to decimal.
Per this online tool:
- $1.111 = 00111111100011100011010100111111$
- $2.222 = 01000000000011100011010100111111$
Summing these two together using signed binary addition, I get:
$0111 1111 1001 1100 0110 1010 0111 1110$
In hexadecimal, this is:
$7F9C6A7E$
And according to this other version of the tool, that corresponds to $NaN$.
What's going on here?
binary floating-point
asked Nov 25 at 0:53
AleksandrH
1,22221123
You can't expect doing integer addition on floating-point representations to give meaningful results.
– Henning Makholm
Nov 25 at 1:01
How would I go about trying to do what I want to do here?
– AleksandrH
Nov 25 at 1:06
I have no idea what it is you want to do. Use floating-point addition rather than integer?
– Henning Makholm
Nov 25 at 1:07
Yes, I was under the impression that once I have the two floating-point numbers represented as binary strings, I could simply add them together bit by bit and then translate the resulting 32-bit string to decimal floating point. The IEEE754 standard defines conversions in both directions (binary to decimal and decimal to binary).
– AleksandrH
Nov 25 at 1:12
You have to adjust them so they have the same mantissa before you add them. You ought to read about what the IEEE754 representation is actually constructed.
– saulspatz
Nov 25 at 1:12
|
show 4 more comments
This isn't for any class or homework. As part of my personal study, I'm trying to better understand the IEEE754 representation of decimal floating-point numbers in binary. I'd like to add two numbers: $1.111$ and $2.222$, then compare the result by converting the IEEE754 representation of the sum back to decimal.
Per this online tool:
- $1.111 = 00111111100011100011010100111111$
- $2.222 = 01000000000011100011010100111111$
Summing these two together using signed binary addition, I get:
$0111 1111 1001 1100 0110 1010 0111 1110$
In hexadecimal, this is:
$7F9C6A7E$
And according to this other version of the tool, that corresponds to $NaN$.
What's going on here?
binary floating-point
asked Nov 25 at 0:53
AleksandrH
1,22221123
This isn't for any class or homework. As part of my personal study, I'm trying to better understand the IEEE754 representation of decimal floating-point numbers in binary. I'd like to add two numbers: $1.111$ and $2.222$, then compare the result by converting the IEEE754 representation of the sum back to decimal.
Per this online tool:
- $1.111 = 00111111100011100011010100111111$
- $2.222 = 01000000000011100011010100111111$
Summing these two together using signed binary addition, I get:
$0111 1111 1001 1100 0110 1010 0111 1110$
In hexadecimal, this is:
$7F9C6A7E$
And according to this other version of the tool, that corresponds to $NaN$.
What's going on here?
binary floating-point
binary floating-point
asked Nov 25 at 0:53
AleksandrH
1,22221123
asked Nov 25 at 0:53
AleksandrH
1,22221123
asked Nov 25 at 0:53
AleksandrH
1,22221123
asked Nov 25 at 0:53
AleksandrH
1,22221123
asked Nov 25 at 0:53
AleksandrH
1,22221123
1,22221123
You can't expect doing integer addition on floating-point representations to give meaningful results.
– Henning Makholm
Nov 25 at 1:01
How would I go about trying to do what I want to do here?
– AleksandrH
Nov 25 at 1:06
I have no idea what it is you want to do. Use floating-point addition rather than integer?
– Henning Makholm
Nov 25 at 1:07
Yes, I was under the impression that once I have the two floating-point numbers represented as binary strings, I could simply add them together bit by bit and then translate the resulting 32-bit string to decimal floating point. The IEEE754 standard defines conversions in both directions (binary to decimal and decimal to binary).
– AleksandrH
Nov 25 at 1:12
You have to adjust them so they have the same mantissa before you add them. You ought to read about what the IEEE754 representation is actually constructed.
– saulspatz
Nov 25 at 1:12
|
show 4 more comments
You can't expect doing integer addition on floating-point representations to give meaningful results.
– Henning Makholm
Nov 25 at 1:01
How would I go about trying to do what I want to do here?
– AleksandrH
Nov 25 at 1:06
I have no idea what it is you want to do. Use floating-point addition rather than integer?
– Henning Makholm
Nov 25 at 1:07
Yes, I was under the impression that once I have the two floating-point numbers represented as binary strings, I could simply add them together bit by bit and then translate the resulting 32-bit string to decimal floating point. The IEEE754 standard defines conversions in both directions (binary to decimal and decimal to binary).
– AleksandrH
Nov 25 at 1:12
You have to adjust them so they have the same mantissa before you add them. You ought to read about what the IEEE754 representation is actually constructed.
– saulspatz
Nov 25 at 1:12
You can't expect doing integer addition on floating-point representations to give meaningful results.
– Henning Makholm
Nov 25 at 1:01
You can't expect doing integer addition on floating-point representations to give meaningful results.
– Henning Makholm
Nov 25 at 1:01
How would I go about trying to do what I want to do here?
– AleksandrH
Nov 25 at 1:06
How would I go about trying to do what I want to do here?
– AleksandrH
Nov 25 at 1:06
I have no idea what it is you want to do. Use floating-point addition rather than integer?
– Henning Makholm
Nov 25 at 1:07
I have no idea what it is you want to do. Use floating-point addition rather than integer?
– Henning Makholm
Nov 25 at 1:07
Yes, I was under the impression that once I have the two floating-point numbers represented as binary strings, I could simply add them together bit by bit and then translate the resulting 32-bit string to decimal floating point. The IEEE754 standard defines conversions in both directions (binary to decimal and decimal to binary).
– AleksandrH
Nov 25 at 1:12
Yes, I was under the impression that once I have the two floating-point numbers represented as binary strings, I could simply add them together bit by bit and then translate the resulting 32-bit string to decimal floating point. The IEEE754 standard defines conversions in both directions (binary to decimal and decimal to binary).
– AleksandrH
Nov 25 at 1:12
You have to adjust them so they have the same mantissa before you add them. You ought to read about what the IEEE754 representation is actually constructed.
– saulspatz
Nov 25 at 1:12
You have to adjust them so they have the same mantissa before you add them. You ought to read about what the IEEE754 representation is actually constructed.
– saulspatz
Nov 25 at 1:12
|
show 4 more comments
1 Answer
1
active
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You cannot expect to use integer binary addition on two floating-point representations and get a meaningful result.
First, $1.111$ cannot be represented exactly in binary floating point. Your 00111111100011100011010100111111 is actually the IEEE-754 single precision representation of the number
$$ 1.11099994182586669921875 $$
which is the closest representable number to $1.111$. This breaks up as
0 01111111 00011100011010100111111
sign biased exponent fractional part of mantissa
and stands for the number
$$ 1.00011100011010100111111_2 times 2^{127-127} $$
The representation of $2.222$ is twice that, with the same mantissa but the exponent one higher. When we add them we must position the mantissas correctly with respect to each other:
1.00011100011010100111111
+ 10.0011100011010100111111
----------------------------
= 11.01010101001111110111101
11.0101010100111111011110 <-- rounded to 1+23 bits mantissa using round-to-even
0 10000000 10101010100111111011110
sign biased exp fractional mantissa
And the representation 01000000010101010100111111011110 corresponds to the number
$$ 3.332999706268310546875 $$
Note that this is not the closest representable number to $3.333$, which would be the next one,
$$ 3.33329999446868896484375 $$
but the round-to-even rule led to rounding down the full result of the addition, which compounded the error inherent in the two inputs each being slightly smaller than $1.111$ and $2.222$.
answered Nov 25 at 1:23
Henning Makholm
238k16303537
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
@AleksandrH: Because the second addend has a biased exponent of10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.
– Henning Makholm
Nov 25 at 1:46
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
add a comment |
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You cannot expect to use integer binary addition on two floating-point representations and get a meaningful result.
First, $1.111$ cannot be represented exactly in binary floating point. Your 00111111100011100011010100111111 is actually the IEEE-754 single precision representation of the number
$$ 1.11099994182586669921875 $$
which is the closest representable number to $1.111$. This breaks up as
0 01111111 00011100011010100111111
sign biased exponent fractional part of mantissa
and stands for the number
$$ 1.00011100011010100111111_2 times 2^{127-127} $$
The representation of $2.222$ is twice that, with the same mantissa but the exponent one higher. When we add them we must position the mantissas correctly with respect to each other:
1.00011100011010100111111
+ 10.0011100011010100111111
----------------------------
= 11.01010101001111110111101
11.0101010100111111011110 <-- rounded to 1+23 bits mantissa using round-to-even
0 10000000 10101010100111111011110
sign biased exp fractional mantissa
And the representation 01000000010101010100111111011110 corresponds to the number
$$ 3.332999706268310546875 $$
Note that this is not the closest representable number to $3.333$, which would be the next one,
$$ 3.33329999446868896484375 $$
but the round-to-even rule led to rounding down the full result of the addition, which compounded the error inherent in the two inputs each being slightly smaller than $1.111$ and $2.222$.
answered Nov 25 at 1:23
Henning Makholm
238k16303537
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
@AleksandrH: Because the second addend has a biased exponent of10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.
– Henning Makholm
Nov 25 at 1:46
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
add a comment |
You cannot expect to use integer binary addition on two floating-point representations and get a meaningful result.
First, $1.111$ cannot be represented exactly in binary floating point. Your 00111111100011100011010100111111 is actually the IEEE-754 single precision representation of the number
$$ 1.11099994182586669921875 $$
which is the closest representable number to $1.111$. This breaks up as
0 01111111 00011100011010100111111
sign biased exponent fractional part of mantissa
and stands for the number
$$ 1.00011100011010100111111_2 times 2^{127-127} $$
The representation of $2.222$ is twice that, with the same mantissa but the exponent one higher. When we add them we must position the mantissas correctly with respect to each other:
1.00011100011010100111111
+ 10.0011100011010100111111
----------------------------
= 11.01010101001111110111101
11.0101010100111111011110 <-- rounded to 1+23 bits mantissa using round-to-even
0 10000000 10101010100111111011110
sign biased exp fractional mantissa
And the representation 01000000010101010100111111011110 corresponds to the number
$$ 3.332999706268310546875 $$
Note that this is not the closest representable number to $3.333$, which would be the next one,
$$ 3.33329999446868896484375 $$
but the round-to-even rule led to rounding down the full result of the addition, which compounded the error inherent in the two inputs each being slightly smaller than $1.111$ and $2.222$.
answered Nov 25 at 1:23
Henning Makholm
238k16303537
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
@AleksandrH: Because the second addend has a biased exponent of10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.
– Henning Makholm
Nov 25 at 1:46
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
add a comment |
You cannot expect to use integer binary addition on two floating-point representations and get a meaningful result.
First, $1.111$ cannot be represented exactly in binary floating point. Your 00111111100011100011010100111111 is actually the IEEE-754 single precision representation of the number
$$ 1.11099994182586669921875 $$
which is the closest representable number to $1.111$. This breaks up as
0 01111111 00011100011010100111111
sign biased exponent fractional part of mantissa
and stands for the number
$$ 1.00011100011010100111111_2 times 2^{127-127} $$
The representation of $2.222$ is twice that, with the same mantissa but the exponent one higher. When we add them we must position the mantissas correctly with respect to each other:
1.00011100011010100111111
+ 10.0011100011010100111111
----------------------------
= 11.01010101001111110111101
11.0101010100111111011110 <-- rounded to 1+23 bits mantissa using round-to-even
0 10000000 10101010100111111011110
sign biased exp fractional mantissa
And the representation 01000000010101010100111111011110 corresponds to the number
$$ 3.332999706268310546875 $$
Note that this is not the closest representable number to $3.333$, which would be the next one,
$$ 3.33329999446868896484375 $$
but the round-to-even rule led to rounding down the full result of the addition, which compounded the error inherent in the two inputs each being slightly smaller than $1.111$ and $2.222$.
answered Nov 25 at 1:23
Henning Makholm
238k16303537
You cannot expect to use integer binary addition on two floating-point representations and get a meaningful result.
First, $1.111$ cannot be represented exactly in binary floating point. Your 00111111100011100011010100111111 is actually the IEEE-754 single precision representation of the number
$$ 1.11099994182586669921875 $$
which is the closest representable number to $1.111$. This breaks up as
0 01111111 00011100011010100111111
sign biased exponent fractional part of mantissa
and stands for the number
$$ 1.00011100011010100111111_2 times 2^{127-127} $$
The representation of $2.222$ is twice that, with the same mantissa but the exponent one higher. When we add them we must position the mantissas correctly with respect to each other:
1.00011100011010100111111
+ 10.0011100011010100111111
----------------------------
= 11.01010101001111110111101
11.0101010100111111011110 <-- rounded to 1+23 bits mantissa using round-to-even
0 10000000 10101010100111111011110
sign biased exp fractional mantissa
And the representation 01000000010101010100111111011110 corresponds to the number
$$ 3.332999706268310546875 $$
Note that this is not the closest representable number to $3.333$, which would be the next one,
$$ 3.33329999446868896484375 $$
but the round-to-even rule led to rounding down the full result of the addition, which compounded the error inherent in the two inputs each being slightly smaller than $1.111$ and $2.222$.
answered Nov 25 at 1:23
Henning Makholm
238k16303537
edited Nov 25 at 1:39
answered Nov 25 at 1:23
Henning Makholm
238k16303537
answered Nov 25 at 1:23
Henning Makholm
238k16303537
answered Nov 25 at 1:23
Henning Makholm
238k16303537
238k16303537
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
@AleksandrH: Because the second addend has a biased exponent of10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.
– Henning Makholm
Nov 25 at 1:46
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
add a comment |
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
@AleksandrH: Because the second addend has a biased exponent of10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.
– Henning Makholm
Nov 25 at 1:46
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
I followed this well until we got to the $10.00...$ part. Why did the decimal point move one place to the right?
– AleksandrH
Nov 25 at 1:43
@AleksandrH: Because the second addend has a biased exponent of
10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.– Henning Makholm
Nov 25 at 1:46
@AleksandrH: Because the second addend has a biased exponent of
10000000, so it represents the number $1.langlemathit{mantissa}rangle_2 times 2^{128-127}$ -- in other words the binary points is shifted one position to the right.– Henning Makholm
Nov 25 at 1:46
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
Yeah, I don't understand. Sorry for wasting your time.
– AleksandrH
Nov 25 at 14:06
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
@AleksandrH: The job of the exponent is to encode where the binary point is. That's what makes the representation "floating point" -- you can move the point! In the $2.22$ representation the exponent is $1$ (after we subtract the fixed bias), meaning that the point is after one of the explicitly represented mantissa bits.
– Henning Makholm
Nov 25 at 14:15
add a comment |
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You can't expect doing integer addition on floating-point representations to give meaningful results.
– Henning Makholm
Nov 25 at 1:01
How would I go about trying to do what I want to do here?
– AleksandrH
Nov 25 at 1:06
I have no idea what it is you want to do. Use floating-point addition rather than integer?
– Henning Makholm
Nov 25 at 1:07
Yes, I was under the impression that once I have the two floating-point numbers represented as binary strings, I could simply add them together bit by bit and then translate the resulting 32-bit string to decimal floating point. The IEEE754 standard defines conversions in both directions (binary to decimal and decimal to binary).
– AleksandrH
Nov 25 at 1:12
You have to adjust them so they have the same mantissa before you add them. You ought to read about what the IEEE754 representation is actually constructed.
– saulspatz
Nov 25 at 1:12