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Unexpected result with right shift after bitwise negation

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17

4

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

share|improve this question

edited Apr 15 at 1:02

John Kugelman

249k54407460

asked Apr 15 at 0:46

IslamIslam

1096

add a comment | 

17

4

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

share|improve this question

edited Apr 15 at 1:02

John Kugelman

249k54407460

asked Apr 15 at 0:46

IslamIslam

1096

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

share|improve this question

edited Apr 15 at 1:02

John Kugelman

249k54407460

asked Apr 15 at 0:46

IslamIslam

1096

int main()

 uint8_t port = 0x5a;
 uint8_t result_8 = (~port) >> 4;
 //result_8 = result_8 >> 4;

 printf("%i", result_8);

 return 0;

share|improve this question

edited Apr 15 at 1:02

John Kugelman

249k54407460

asked Apr 15 at 0:46

IslamIslam

1096

share|improve this question

edited Apr 15 at 1:02

John Kugelman

249k54407460

asked Apr 15 at 0:46

IslamIslam

1096

edited Apr 15 at 1:02

John Kugelman

249k54407460

edited Apr 15 at 1:02

John Kugelman

249k54407460

asked Apr 15 at 0:46

IslamIslam

1096

asked Apr 15 at 0:46

IslamIslam

1096

1 Answer
1

active

oldest

votes

26

C promotes uint8_t to int before doing operations on it. So:

1. 
   port is promoted to signed integer 0x0000005a.


2. 
   ~ inverts it giving 0xffffffa5.


3. An arithmetic shift returns 0xfffffffa.


4. It's truncated back into a uint8_t giving 0xfa == 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;

share|improve this answer

edited Apr 15 at 5:37

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  


• 
  
  
  

  
  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
  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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26

C promotes uint8_t to int before doing operations on it. So:

1. 
   port is promoted to signed integer 0x0000005a.


2. 
   ~ inverts it giving 0xffffffa5.


3. An arithmetic shift returns 0xfffffffa.


4. It's truncated back into a uint8_t giving 0xfa == 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;

share|improve this answer

edited Apr 15 at 5:37

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  


• 
  
  
  

  
  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
  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

26

C promotes uint8_t to int before doing operations on it. So:

1. 
   port is promoted to signed integer 0x0000005a.


2. 
   ~ inverts it giving 0xffffffa5.


3. An arithmetic shift returns 0xfffffffa.


4. It's truncated back into a uint8_t giving 0xfa == 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;

share|improve this answer

edited Apr 15 at 5:37

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  


• 
  
  
  

  
  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
  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:

1. 
   port is promoted to signed integer 0x0000005a.


2. 
   ~ inverts it giving 0xffffffa5.


3. An arithmetic shift returns 0xfffffffa.


4. It's truncated back into a uint8_t giving 0xfa == 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;

share|improve this answer

edited Apr 15 at 5:37

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163

uint8_t result_8 = (uint8_t)(~port) >> 4;

uint8_t result_8 = (~port & 0xff) >> 4;

uint8_t result_8 = (port ^ 0xff) >> 4;

share|improve this answer

edited Apr 15 at 5:37

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163

answered Apr 15 at 0:50

ybungalobillybungalobill

46.6k1397163