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Unexpected result with right shift after bitwise negation



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Should we burninate the [wrap] tag?What 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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16















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































    16















    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



























      16












      16








      16


      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 yesterday









      John Kugelman

      249k54407460




      249k54407460










      asked yesterday









      IslamIslam

      1046




      1046
























          1 Answer
          1






          active

          oldest

          votes


















          24














          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


























          • 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





            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





            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














          Your Answer






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          1 Answer
          1






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          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          24














          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


























          • 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





            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





            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


















          24














          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


























          • 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





            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





            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
















          24












          24








          24







          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















          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














          share|improve this answer



          share|improve this answer








          edited 22 hours ago

























          answered yesterday









          ybungalobillybungalobill

          46.3k1396163




          46.3k1396163













          • 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





            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





            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






          • 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






          • 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






















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