Unexpected result with right shift after bitwise negation





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7















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































    7















    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



























      7












      7








      7


      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
















      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      John Kugelman

      249k54407460




      249k54407460










      asked 2 hours ago









      IslamIslam

      595




      595
























          1 Answer
          1






          active

          oldest

          votes


















          11














          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;


          or mask it:



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





          share|improve this answer
























          • 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
            1 hour ago






          • 3





            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
            1 hour ago






          • 1





            Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

            – Nayuki
            42 mins ago











          • @Nayuki: that's a good one too!

            – ybungalobill
            33 mins ago












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          1 Answer
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          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          11














          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;


          or mask it:



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





          share|improve this answer
























          • 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
            1 hour ago






          • 3





            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
            1 hour ago






          • 1





            Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

            – Nayuki
            42 mins ago











          • @Nayuki: that's a good one too!

            – ybungalobill
            33 mins ago
















          11














          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;


          or mask it:



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





          share|improve this answer
























          • 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
            1 hour ago






          • 3





            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
            1 hour ago






          • 1





            Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

            – Nayuki
            42 mins ago











          • @Nayuki: that's a good one too!

            – ybungalobill
            33 mins ago














          11












          11








          11







          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;


          or mask it:



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





          share|improve this answer













          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;


          or mask it:



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






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 2 hours ago









          ybungalobillybungalobill

          46.2k1396163




          46.2k1396163













          • 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
            1 hour ago






          • 3





            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
            1 hour ago






          • 1





            Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

            – Nayuki
            42 mins ago











          • @Nayuki: that's a good one too!

            – ybungalobill
            33 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
            1 hour ago






          • 3





            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
            1 hour ago






          • 1





            Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

            – Nayuki
            42 mins ago











          • @Nayuki: that's a good one too!

            – ybungalobill
            33 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
          1 hour 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
          1 hour ago




          3




          3





          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
          1 hour 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
          1 hour ago




          1




          1





          Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

          – Nayuki
          42 mins ago





          Or explicitly only flip the low 8 bits: result = (port ^ 0xFF) >> 4;

          – Nayuki
          42 mins ago













          @Nayuki: that's a good one too!

          – ybungalobill
          33 mins ago





          @Nayuki: that's a good one too!

          – ybungalobill
          33 mins ago




















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