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Unexpected result with right shift after bitwise negation
What 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
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
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
254k56413465
asked Apr 15 at 0:46
IslamIslam
1296
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 Apr 15 at 1:02
John Kugelman
254k56413465
asked Apr 15 at 0:46
IslamIslam
1296
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 Apr 15 at 1:02
John Kugelman
254k56413465
asked Apr 15 at 0:46
IslamIslam
1296
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
254k56413465
asked Apr 15 at 0:46
IslamIslam
1296
edited Apr 15 at 1:02
John Kugelman
254k56413465
asked Apr 15 at 0:46
IslamIslam
1296
edited Apr 15 at 1:02
John Kugelman
254k56413465
edited Apr 15 at 1:02
John Kugelman
254k56413465
edited Apr 15 at 1:02
John Kugelman
254k56413465
254k56413465
asked Apr 15 at 0:46
IslamIslam
1296
asked Apr 15 at 0:46
IslamIslam
1296
asked Apr 15 at 0:46
IslamIslam
1296
1296
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;
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
47.9k1399166
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
13
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
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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;
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
47.9k1399166
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
13
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
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
47.9k1399166
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
13
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
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
47.9k1399166
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
47.9k1399166
edited Apr 15 at 5:37
answered Apr 15 at 0:50
ybungalobillybungalobill
47.9k1399166
answered Apr 15 at 0:50
ybungalobillybungalobill
47.9k1399166
answered Apr 15 at 0:50
ybungalobillybungalobill
47.9k1399166
47.9k1399166
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
13
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
13
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
13
13
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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