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
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 19 mins ago
John Kugelman
249k54407460
249k54407460
asked 35 mins 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:
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;
or mask it:
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
17 mins ago
2
uint8_t
is, very likely, a synonym ofunsigned char
on your system. The promotion rules apply to all integral types smaller thanint
.
– ybungalobill
14 mins ago
add a comment |
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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;
or mask it:
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
17 mins ago
2
uint8_t
is, very likely, a synonym ofunsigned char
on your system. The promotion rules apply to all integral types smaller thanint
.
– ybungalobill
14 mins ago
add a comment |
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;
or mask it:
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
17 mins ago
2
uint8_t
is, very likely, a synonym ofunsigned char
on your system. The promotion rules apply to all integral types smaller thanint
.
– ybungalobill
14 mins ago
add a comment |
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;
or mask it:
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;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 32 mins ago
ybungalobillybungalobill
46.1k1395162
46.1k1395162
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
17 mins ago
2
uint8_t
is, very likely, a synonym ofunsigned char
on your system. The promotion rules apply to all integral types smaller thanint
.
– ybungalobill
14 mins ago
add a comment |
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
17 mins ago
2
uint8_t
is, very likely, a synonym ofunsigned char
on your system. The promotion rules apply to all integral types smaller thanint
.
– ybungalobill
14 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
17 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
17 mins ago
2
2
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
14 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
14 mins ago
add a comment |
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