Unexpected result with right shift after bitwise negation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) The Ask Question Wizard is Live! Data science time! April 2019 and salary with experienceWhat are bitwise shift (bit-shift) operators and how do they work?Improve INSERT-per-second performance of SQLite?Right shift two's complement number like an unsigned intbit shifting in C, unexpected resultRight shift with zeros at the beginningUnexepected behavior from multiple bitwise shifts on the same lineUnexpected Result After Arithmetically Right ShiftingWhy unsigned int right shift is always filled with '1'Unusual behavior with shift-right bitwise operatorprintf() function in loop #3 gives unexpected result
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Unexpected result with right shift after bitwise negation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experienceWhat are bitwise shift (bit-shift) operators and how do they work?Improve INSERT-per-second performance of SQLite?Right shift two's complement number like an unsigned intbit shifting in C, unexpected resultRight shift with zeros at the beginningUnexepected behavior from multiple bitwise shifts on the same lineUnexpected Result After Arithmetically Right ShiftingWhy unsigned int right shift is always filled with '1'Unusual behavior with shift-right bitwise operatorprintf() function in loop #3 gives unexpected result
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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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
edited 1 hour ago
John Kugelman
249k54407460
asked 1 hour ago
IslamIslam
545
edited 1 hour ago
John Kugelman
249k54407460
edited 1 hour ago
John Kugelman
249k54407460
edited 1 hour 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
46.2k1396163
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
1 hour ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
1 hour ago
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
8 mins ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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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
46.2k1396163
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
1 hour ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
1 hour ago
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
8 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
46.2k1396163
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
1 hour ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
1 hour ago
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
8 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
46.2k1396163
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
46.2k1396163
answered 1 hour ago
ybungalobillybungalobill
46.2k1396163
answered 1 hour ago
ybungalobillybungalobill
46.2k1396163
answered 1 hour ago
ybungalobillybungalobill
46.2k1396163
46.2k1396163
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
1 hour ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
1 hour ago
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
8 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
1 hour ago
3
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
1 hour ago
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
8 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Or explicitly only flip the low 8 bits:
result = (port ^ 0xFF) >> 4;– Nayuki
8 mins ago
Or explicitly only flip the low 8 bits:
result = (port ^ 0xFF) >> 4;– Nayuki
8 mins ago
add a comment |
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