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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
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
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
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
249k54407460
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:
port
is promoted to signed integer0x0000005a
.
~
inverts it giving0xffffffa5
.- An arithmetic shift returns
0xfffffffa
. - It's truncated back into a
uint8_t
giving0xfa == 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;
you're right but i think C doesn't promote onlyuint8_t
but alsounsigned char
because i tested it withunsigned char
too and got the same result! Am i right?
– Islam
yesterday
11
uint8_t
is, very likely, a synonym ofunsigned char
on 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
1
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:
port
is promoted to signed integer0x0000005a
.
~
inverts it giving0xffffffa5
.- An arithmetic shift returns
0xfffffffa
. - It's truncated back into a
uint8_t
giving0xfa == 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;
you're right but i think C doesn't promote onlyuint8_t
but alsounsigned char
because i tested it withunsigned char
too and got the same result! Am i right?
– Islam
yesterday
11
uint8_t
is, very likely, a synonym ofunsigned char
on 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:
port
is promoted to signed integer0x0000005a
.
~
inverts it giving0xffffffa5
.- An arithmetic shift returns
0xfffffffa
. - It's truncated back into a
uint8_t
giving0xfa == 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;
you're right but i think C doesn't promote onlyuint8_t
but alsounsigned char
because i tested it withunsigned char
too and got the same result! Am i right?
– Islam
yesterday
11
uint8_t
is, very likely, a synonym ofunsigned char
on 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:
port
is promoted to signed integer0x0000005a
.
~
inverts it giving0xffffffa5
.- An arithmetic shift returns
0xfffffffa
. - It's truncated back into a
uint8_t
giving0xfa == 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;
C promotes uint8_t
to int
before doing operations on it. So:
port
is promoted to signed integer0x0000005a
.
~
inverts it giving0xffffffa5
.- An arithmetic shift returns
0xfffffffa
. - It's truncated back into a
uint8_t
giving0xfa == 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;
edited 22 hours ago
answered yesterday
ybungalobillybungalobill
46.3k1396163
46.3k1396163
you're right but i think C doesn't promote onlyuint8_t
but alsounsigned char
because i tested it withunsigned char
too and got the same result! Am i right?
– Islam
yesterday
11
uint8_t
is, very likely, a synonym ofunsigned char
on 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_t
but alsounsigned char
because i tested it withunsigned char
too and got the same result! Am i right?
– Islam
yesterday
11
uint8_t
is, very likely, a synonym ofunsigned char
on 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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