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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6















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;










share|improve this question






























    6















    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;










    share|improve this question


























      6












      6








      6


      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;










      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 19 mins ago









      John Kugelman

      249k54407460




      249k54407460










      asked 35 mins ago









      IslamIslam

      545




      545






















          1 Answer
          1






          active

          oldest

          votes


















          9














          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
            17 mins ago






          • 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











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          1 Answer
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          active

          oldest

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          active

          oldest

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          9














          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
            17 mins ago






          • 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















          9














          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
            17 mins ago






          • 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













          9












          9








          9







          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 32 mins ago









          ybungalobillybungalobill

          46.1k1395162




          46.1k1395162












          • 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





            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

















          • 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





            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
















          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



















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