Unexpected result with right shift after bitwise negation
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18
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
edited Apr 15 at 1:02
John Kugelman
249k54407460
asked Apr 15 at 0:46
IslamIslam
1146
add a comment |
18
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
edited Apr 15 at 1:02
John Kugelman
249k54407460
asked Apr 15 at 0:46
IslamIslam
1146
add a comment |
18
18
18
4
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
edited Apr 15 at 1:02
John Kugelman
249k54407460
asked Apr 15 at 0:46
IslamIslam
1146
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
c bit-manipulation
edited Apr 15 at 1:02
John Kugelman
249k54407460
asked Apr 15 at 0:46
IslamIslam
1146
edited Apr 15 at 1:02
John Kugelman
249k54407460
asked Apr 15 at 0:46
IslamIslam
1146
edited Apr 15 at 1:02
John Kugelman
249k54407460
edited Apr 15 at 1:02
John Kugelman
249k54407460
edited Apr 15 at 1:02
John Kugelman
249k54407460
249k54407460
asked Apr 15 at 0:46
IslamIslam
1146
asked Apr 15 at 0:46
IslamIslam
1146
asked Apr 15 at 0:46
IslamIslam
1146
1146
add a comment |
add a comment |
1 Answer
1
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oldest
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27
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4;
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
Apr 15 at 1:04
12
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
Apr 15 at 1:07
7
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
Apr 15 at 2:16
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
|
show 2 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
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27
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4;
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
Apr 15 at 1:04
12
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
Apr 15 at 1:07
7
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
Apr 15 at 2:16
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
|
show 2 more comments
27
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4;
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
Apr 15 at 1:04
12
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
Apr 15 at 1:07
7
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
Apr 15 at 2:16
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
|
show 2 more comments
27
27
27
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4;
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4;
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
edited Apr 15 at 5:37
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
answered Apr 15 at 0:50
ybungalobillybungalobill
46.6k1397163
46.6k1397163
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
Apr 15 at 1:04
12
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
Apr 15 at 1:07
7
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
Apr 15 at 2:16
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
|
show 2 more comments
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
Apr 15 at 1:04
12
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
Apr 15 at 1:07
7
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
Apr 15 at 2:16
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
you're right but i think C doesn't promote only
uint8_t but also unsigned char because i tested it with unsigned char too and got the same result! Am i right?– Islam
Apr 15 at 1:04
you're right but i think C doesn't promote only
uint8_t but also unsigned char because i tested it with unsigned char too and got the same result! Am i right?– Islam
Apr 15 at 1:04
12
12
uint8_t is, very likely, a synonym of unsigned char on your system. The promotion rules apply to all integral types smaller than int.– ybungalobill
Apr 15 at 1:07
uint8_t is, very likely, a synonym of unsigned char on your system. The promotion rules apply to all integral types smaller than int.– ybungalobill
Apr 15 at 1:07
7
7
Or explicitly only flip the low 8 bits:
result = (port ^ 0xFF) >> 4;– Nayuki
Apr 15 at 2:16
Or explicitly only flip the low 8 bits:
result = (port ^ 0xFF) >> 4;– Nayuki
Apr 15 at 2:16
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
@Nayuki: that's a good one too!
– ybungalobill
Apr 15 at 2:26
2
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
Apr 15 at 9:02
|
show 2 more comments
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