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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 yesterday
John Kugelman
249k54407460
asked yesterday
IslamIslam
1046
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 yesterday
John Kugelman
249k54407460
asked yesterday
IslamIslam
1046
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 yesterday
John Kugelman
249k54407460
asked yesterday
IslamIslam
1046
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 yesterday
John Kugelman
249k54407460
asked yesterday
IslamIslam
1046
edited yesterday
John Kugelman
249k54407460
asked yesterday
IslamIslam
1046
edited yesterday
John Kugelman
249k54407460
edited yesterday
John Kugelman
249k54407460
edited yesterday
John Kugelman
249k54407460
249k54407460
asked yesterday
IslamIslam
1046
asked yesterday
IslamIslam
1046
asked yesterday
IslamIslam
1046
1046
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 yesterday
ybungalobillybungalobill
46.3k1396163
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
yesterday
11
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
yesterday
6
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours ago
|
show 2 more comments
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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;
mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4;
answered yesterday
ybungalobillybungalobill
46.3k1396163
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
yesterday
11
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
yesterday
6
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours ago
|
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 yesterday
ybungalobillybungalobill
46.3k1396163
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
yesterday
11
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
yesterday
6
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours ago
|
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 yesterday
ybungalobillybungalobill
46.3k1396163
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 yesterday
ybungalobillybungalobill
46.3k1396163
edited 22 hours ago
answered yesterday
ybungalobillybungalobill
46.3k1396163
answered yesterday
ybungalobillybungalobill
46.3k1396163
answered yesterday
ybungalobillybungalobill
46.3k1396163
46.3k1396163
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
yesterday
11
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
yesterday
6
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours ago
|
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
yesterday
11
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
yesterday
6
Or explicitly only flip the low 8 bits:result = (port ^ 0xFF) >> 4;
– Nayuki
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours 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
yesterday
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
yesterday
11
11
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
yesterday
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
yesterday
6
6
Or explicitly only flip the low 8 bits:
result = (port ^ 0xFF) >> 4;– Nayuki
yesterday
Or explicitly only flip the low 8 bits:
result = (port ^ 0xFF) >> 4;– Nayuki
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
@Nayuki: that's a good one too!
– ybungalobill
yesterday
2
2
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours ago
@TomášZato: Yes. See en.cppreference.com/w/cpp/language/…
– ybungalobill
18 hours ago
|
show 2 more comments
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