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Unexpected result with right shift after bitwise negation
Unexpected result with right shift after bitwise negation
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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 53 mins ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
add a comment |
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 53 mins ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
add a comment |
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 53 mins ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
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 53 mins ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
edited 53 mins ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
edited 53 mins ago
John Kugelman
249k54407460
edited 53 mins ago
John Kugelman
249k54407460
edited 53 mins ago
John Kugelman
249k54407460
249k54407460
asked 1 hour ago
IslamIslam
545
asked 1 hour ago
IslamIslam
545
asked 1 hour ago
IslamIslam
545
545
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
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
51 mins ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
48 mins ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
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
51 mins ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
48 mins ago
add a comment |
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;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
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
51 mins ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
48 mins ago
add a comment |
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;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
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;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
answered 1 hour ago
ybungalobillybungalobill
46.1k1396163
46.1k1396163
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
51 mins ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
48 mins ago
add a comment |
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
51 mins ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
48 mins ago
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
51 mins ago
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
51 mins ago
3
3
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
48 mins ago
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
48 mins ago
add a comment |
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