Change Arithmetic Right Shift to Logical Right Shift












0












$begingroup$


The following code is a solution to a textbook (Bryant&O'Hallaron: Computer Systems A programmer's Perspective 2nd Ed) problem in bit-level data manipulation (attempted for the challenge, not a class). The function srl is not written necessarily in a practical manner, but within the constraints required by the problem.



Questions



Is there is a clearer, more straight-forward way to write this within the required constraints of the problem (perhaps with fewer ~ operations)?



There is a need to avoid undefined behavior of the left shift, when k = 0. In this case int_bits - k = int_bits, which causes the shift to work unpredictably. Is there a better way to handle the undefined behavior of the shift operations, when the shift is larger than the number of bits in the interger?



It seems to work correctly, but I lack an answer, so any feedback on the solution would be appreciated.



Requirements





  • No additional right shifts or type casts may be used beyond the given expression



      /*Perform shift arithmetically*/
    unsigned xsra = (int) x >> k;


  • Only addition or subtraction may be used, no multiplication, division, or modulus


  • No comparison operators and no conditionals


  • Only bit-wise operations (except further right shifts) and logical operators may be used



Code



unsigned srl(unsigned x, int k) {
/*Perform shift arithmetically*/
unsigned xsra = (int) x >> k;

unsigned int_bits = sizeof(int) << 3;//calculates the number of bits in int (assumes 8-bit byte)

unsigned zero_or_all_bits = ~0 + !k;//for k = 0, corrects for the undefined behavior in
//the left shift produced from int_bits - k = int_bits
//if k != 0, !k == 0, and zero_or_all_bits == ~0
//if k == 0, zero_or_all_bits == 0

unsigned high_bit_mask = ~(zero_or_all_bits << (zero_or_all_bits & (int_bits - k)));
/******************************************/
//creates a mask of either all bits set in an unsigned int (UINT_MAX)
//or a mask with k high bits cleared.
//if k == 0, then high_bit_mask = ~(0 << 0) = ~0.
//if k != 0, then high_bit_mask = ~(~0 << (~0 & (int_bits - k)))
//ex. k == 3, high_bit_mask == 0x1FFFFFFF
//ex. k == 0, high_bit_mask == 0xFFFFFFFF
//ex. k == 31, high_bit_mask == 0xFFFFFFFE
/******************************************/

return xsra & high_bit_mask;


}





Test Code



printf("Test srl:n");
printf("srl(-1, 1): 0x%.8xn", srl(-1, 1));
printf("srl(-1, 4): 0x%.8xn", srl(-1, 4));
printf("srl(-1, 5): 0x%.8xn", srl(-1, 5));
printf("srl(-1, 31): 0x%.8xn", srl(-1, 31));
printf("srl(-1, 0): 0x%.8xn", srl(-1, 0));

printf("srl(0x7FFFFFFF, 1): 0x%.8xn", srl(0x7FFFFFFF, 1));
printf("srl(0x7FFFFFFF, 4): 0x%.8xn", srl(0x7FFFFFFF, 4));
printf("srl(0x7FFFFFFF, 5): 0x%.8xn", srl(0x7FFFFFFF, 5));
printf("srl(0x7FFFFFFF, 31): 0x%.8xn", srl(0x7FFFFFFF, 31));
printf("srl(0x7FFFFFFF, 0): 0x%.8xn", srl(0x7FFFFFFF, 0));

printf("srl(0x80000000, 1): 0x%.8xn", srl(0x80000000, 1));
printf("srl(0x80000000, 4): 0x%.8xn", srl(0x80000000, 4));
printf("srl(0x80000000, 5): 0x%.8xn", srl(0x80000000, 5));
printf("srl(0x80000000, 31): 0x%.8xn", srl(0x80000000, 31));
printf("srl(0x80000000, 0): 0x%.8xn", srl(0x80000000, 0));

printf("srl(0, 1): 0x%.8xn", srl(0, 1));
printf("srl(0, 4): 0x%.8xn", srl(0, 4));
printf("srl(0, 5): 0x%.8xn", srl(0, 5));
printf("srl(0, 31): 0x%.8xn", srl(0, 31));
printf("srl(0, 0): 0x%.8xn", srl(0, 0));

printf("srl(1, 1): 0x%.8xn", srl(1, 1));
printf("srl(1, 4): 0x%.8xn", srl(1, 4));
printf("srl(1, 5): 0x%.8xn", srl(1, 5));
printf("srl(1, 31): 0x%.8xn", srl(1, 31));
printf("srl(1, 0): 0x%.8xn", srl(1, 0));




Output




Test srl:

srl(-1, 1): 0x7fffffff

srl(-1, 4): 0x0fffffff

srl(-1, 5): 0x07ffffff

srl(-1, 31): 0x00000001

srl(-1, 0): 0xffffffff

srl(0x7FFFFFFF, 1): 0x3fffffff

srl(0x7FFFFFFF, 4): 0x07ffffff

srl(0x7FFFFFFF, 5): 0x03ffffff

srl(0x7FFFFFFF, 31): 0x00000000

srl(0x7FFFFFFF, 0): 0x7fffffff

srl(0x80000000, 1): 0x40000000

srl(0x80000000, 4): 0x08000000

srl(0x80000000, 5): 0x04000000

srl(0x80000000, 31): 0x00000001

srl(0x80000000, 0): 0x80000000

srl(0, 1): 0x00000000

srl(0, 4): 0x00000000

srl(0, 5): 0x00000000

srl(0, 31): 0x00000000

srl(0, 0): 0x00000000

srl(1, 1): 0x00000000

srl(1, 4): 0x00000000

srl(1, 5): 0x00000000

srl(1, 31): 0x00000000

srl(1, 0): 0x00000001











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$endgroup$

















    0












    $begingroup$


    The following code is a solution to a textbook (Bryant&O'Hallaron: Computer Systems A programmer's Perspective 2nd Ed) problem in bit-level data manipulation (attempted for the challenge, not a class). The function srl is not written necessarily in a practical manner, but within the constraints required by the problem.



    Questions



    Is there is a clearer, more straight-forward way to write this within the required constraints of the problem (perhaps with fewer ~ operations)?



    There is a need to avoid undefined behavior of the left shift, when k = 0. In this case int_bits - k = int_bits, which causes the shift to work unpredictably. Is there a better way to handle the undefined behavior of the shift operations, when the shift is larger than the number of bits in the interger?



    It seems to work correctly, but I lack an answer, so any feedback on the solution would be appreciated.



    Requirements





    • No additional right shifts or type casts may be used beyond the given expression



        /*Perform shift arithmetically*/
      unsigned xsra = (int) x >> k;


    • Only addition or subtraction may be used, no multiplication, division, or modulus


    • No comparison operators and no conditionals


    • Only bit-wise operations (except further right shifts) and logical operators may be used



    Code



    unsigned srl(unsigned x, int k) {
    /*Perform shift arithmetically*/
    unsigned xsra = (int) x >> k;

    unsigned int_bits = sizeof(int) << 3;//calculates the number of bits in int (assumes 8-bit byte)

    unsigned zero_or_all_bits = ~0 + !k;//for k = 0, corrects for the undefined behavior in
    //the left shift produced from int_bits - k = int_bits
    //if k != 0, !k == 0, and zero_or_all_bits == ~0
    //if k == 0, zero_or_all_bits == 0

    unsigned high_bit_mask = ~(zero_or_all_bits << (zero_or_all_bits & (int_bits - k)));
    /******************************************/
    //creates a mask of either all bits set in an unsigned int (UINT_MAX)
    //or a mask with k high bits cleared.
    //if k == 0, then high_bit_mask = ~(0 << 0) = ~0.
    //if k != 0, then high_bit_mask = ~(~0 << (~0 & (int_bits - k)))
    //ex. k == 3, high_bit_mask == 0x1FFFFFFF
    //ex. k == 0, high_bit_mask == 0xFFFFFFFF
    //ex. k == 31, high_bit_mask == 0xFFFFFFFE
    /******************************************/

    return xsra & high_bit_mask;


    }





    Test Code



    printf("Test srl:n");
    printf("srl(-1, 1): 0x%.8xn", srl(-1, 1));
    printf("srl(-1, 4): 0x%.8xn", srl(-1, 4));
    printf("srl(-1, 5): 0x%.8xn", srl(-1, 5));
    printf("srl(-1, 31): 0x%.8xn", srl(-1, 31));
    printf("srl(-1, 0): 0x%.8xn", srl(-1, 0));

    printf("srl(0x7FFFFFFF, 1): 0x%.8xn", srl(0x7FFFFFFF, 1));
    printf("srl(0x7FFFFFFF, 4): 0x%.8xn", srl(0x7FFFFFFF, 4));
    printf("srl(0x7FFFFFFF, 5): 0x%.8xn", srl(0x7FFFFFFF, 5));
    printf("srl(0x7FFFFFFF, 31): 0x%.8xn", srl(0x7FFFFFFF, 31));
    printf("srl(0x7FFFFFFF, 0): 0x%.8xn", srl(0x7FFFFFFF, 0));

    printf("srl(0x80000000, 1): 0x%.8xn", srl(0x80000000, 1));
    printf("srl(0x80000000, 4): 0x%.8xn", srl(0x80000000, 4));
    printf("srl(0x80000000, 5): 0x%.8xn", srl(0x80000000, 5));
    printf("srl(0x80000000, 31): 0x%.8xn", srl(0x80000000, 31));
    printf("srl(0x80000000, 0): 0x%.8xn", srl(0x80000000, 0));

    printf("srl(0, 1): 0x%.8xn", srl(0, 1));
    printf("srl(0, 4): 0x%.8xn", srl(0, 4));
    printf("srl(0, 5): 0x%.8xn", srl(0, 5));
    printf("srl(0, 31): 0x%.8xn", srl(0, 31));
    printf("srl(0, 0): 0x%.8xn", srl(0, 0));

    printf("srl(1, 1): 0x%.8xn", srl(1, 1));
    printf("srl(1, 4): 0x%.8xn", srl(1, 4));
    printf("srl(1, 5): 0x%.8xn", srl(1, 5));
    printf("srl(1, 31): 0x%.8xn", srl(1, 31));
    printf("srl(1, 0): 0x%.8xn", srl(1, 0));




    Output




    Test srl:

    srl(-1, 1): 0x7fffffff

    srl(-1, 4): 0x0fffffff

    srl(-1, 5): 0x07ffffff

    srl(-1, 31): 0x00000001

    srl(-1, 0): 0xffffffff

    srl(0x7FFFFFFF, 1): 0x3fffffff

    srl(0x7FFFFFFF, 4): 0x07ffffff

    srl(0x7FFFFFFF, 5): 0x03ffffff

    srl(0x7FFFFFFF, 31): 0x00000000

    srl(0x7FFFFFFF, 0): 0x7fffffff

    srl(0x80000000, 1): 0x40000000

    srl(0x80000000, 4): 0x08000000

    srl(0x80000000, 5): 0x04000000

    srl(0x80000000, 31): 0x00000001

    srl(0x80000000, 0): 0x80000000

    srl(0, 1): 0x00000000

    srl(0, 4): 0x00000000

    srl(0, 5): 0x00000000

    srl(0, 31): 0x00000000

    srl(0, 0): 0x00000000

    srl(1, 1): 0x00000000

    srl(1, 4): 0x00000000

    srl(1, 5): 0x00000000

    srl(1, 31): 0x00000000

    srl(1, 0): 0x00000001











    share|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The following code is a solution to a textbook (Bryant&O'Hallaron: Computer Systems A programmer's Perspective 2nd Ed) problem in bit-level data manipulation (attempted for the challenge, not a class). The function srl is not written necessarily in a practical manner, but within the constraints required by the problem.



      Questions



      Is there is a clearer, more straight-forward way to write this within the required constraints of the problem (perhaps with fewer ~ operations)?



      There is a need to avoid undefined behavior of the left shift, when k = 0. In this case int_bits - k = int_bits, which causes the shift to work unpredictably. Is there a better way to handle the undefined behavior of the shift operations, when the shift is larger than the number of bits in the interger?



      It seems to work correctly, but I lack an answer, so any feedback on the solution would be appreciated.



      Requirements





      • No additional right shifts or type casts may be used beyond the given expression



          /*Perform shift arithmetically*/
        unsigned xsra = (int) x >> k;


      • Only addition or subtraction may be used, no multiplication, division, or modulus


      • No comparison operators and no conditionals


      • Only bit-wise operations (except further right shifts) and logical operators may be used



      Code



      unsigned srl(unsigned x, int k) {
      /*Perform shift arithmetically*/
      unsigned xsra = (int) x >> k;

      unsigned int_bits = sizeof(int) << 3;//calculates the number of bits in int (assumes 8-bit byte)

      unsigned zero_or_all_bits = ~0 + !k;//for k = 0, corrects for the undefined behavior in
      //the left shift produced from int_bits - k = int_bits
      //if k != 0, !k == 0, and zero_or_all_bits == ~0
      //if k == 0, zero_or_all_bits == 0

      unsigned high_bit_mask = ~(zero_or_all_bits << (zero_or_all_bits & (int_bits - k)));
      /******************************************/
      //creates a mask of either all bits set in an unsigned int (UINT_MAX)
      //or a mask with k high bits cleared.
      //if k == 0, then high_bit_mask = ~(0 << 0) = ~0.
      //if k != 0, then high_bit_mask = ~(~0 << (~0 & (int_bits - k)))
      //ex. k == 3, high_bit_mask == 0x1FFFFFFF
      //ex. k == 0, high_bit_mask == 0xFFFFFFFF
      //ex. k == 31, high_bit_mask == 0xFFFFFFFE
      /******************************************/

      return xsra & high_bit_mask;


      }





      Test Code



      printf("Test srl:n");
      printf("srl(-1, 1): 0x%.8xn", srl(-1, 1));
      printf("srl(-1, 4): 0x%.8xn", srl(-1, 4));
      printf("srl(-1, 5): 0x%.8xn", srl(-1, 5));
      printf("srl(-1, 31): 0x%.8xn", srl(-1, 31));
      printf("srl(-1, 0): 0x%.8xn", srl(-1, 0));

      printf("srl(0x7FFFFFFF, 1): 0x%.8xn", srl(0x7FFFFFFF, 1));
      printf("srl(0x7FFFFFFF, 4): 0x%.8xn", srl(0x7FFFFFFF, 4));
      printf("srl(0x7FFFFFFF, 5): 0x%.8xn", srl(0x7FFFFFFF, 5));
      printf("srl(0x7FFFFFFF, 31): 0x%.8xn", srl(0x7FFFFFFF, 31));
      printf("srl(0x7FFFFFFF, 0): 0x%.8xn", srl(0x7FFFFFFF, 0));

      printf("srl(0x80000000, 1): 0x%.8xn", srl(0x80000000, 1));
      printf("srl(0x80000000, 4): 0x%.8xn", srl(0x80000000, 4));
      printf("srl(0x80000000, 5): 0x%.8xn", srl(0x80000000, 5));
      printf("srl(0x80000000, 31): 0x%.8xn", srl(0x80000000, 31));
      printf("srl(0x80000000, 0): 0x%.8xn", srl(0x80000000, 0));

      printf("srl(0, 1): 0x%.8xn", srl(0, 1));
      printf("srl(0, 4): 0x%.8xn", srl(0, 4));
      printf("srl(0, 5): 0x%.8xn", srl(0, 5));
      printf("srl(0, 31): 0x%.8xn", srl(0, 31));
      printf("srl(0, 0): 0x%.8xn", srl(0, 0));

      printf("srl(1, 1): 0x%.8xn", srl(1, 1));
      printf("srl(1, 4): 0x%.8xn", srl(1, 4));
      printf("srl(1, 5): 0x%.8xn", srl(1, 5));
      printf("srl(1, 31): 0x%.8xn", srl(1, 31));
      printf("srl(1, 0): 0x%.8xn", srl(1, 0));




      Output




      Test srl:

      srl(-1, 1): 0x7fffffff

      srl(-1, 4): 0x0fffffff

      srl(-1, 5): 0x07ffffff

      srl(-1, 31): 0x00000001

      srl(-1, 0): 0xffffffff

      srl(0x7FFFFFFF, 1): 0x3fffffff

      srl(0x7FFFFFFF, 4): 0x07ffffff

      srl(0x7FFFFFFF, 5): 0x03ffffff

      srl(0x7FFFFFFF, 31): 0x00000000

      srl(0x7FFFFFFF, 0): 0x7fffffff

      srl(0x80000000, 1): 0x40000000

      srl(0x80000000, 4): 0x08000000

      srl(0x80000000, 5): 0x04000000

      srl(0x80000000, 31): 0x00000001

      srl(0x80000000, 0): 0x80000000

      srl(0, 1): 0x00000000

      srl(0, 4): 0x00000000

      srl(0, 5): 0x00000000

      srl(0, 31): 0x00000000

      srl(0, 0): 0x00000000

      srl(1, 1): 0x00000000

      srl(1, 4): 0x00000000

      srl(1, 5): 0x00000000

      srl(1, 31): 0x00000000

      srl(1, 0): 0x00000001











      share|improve this question









      $endgroup$




      The following code is a solution to a textbook (Bryant&O'Hallaron: Computer Systems A programmer's Perspective 2nd Ed) problem in bit-level data manipulation (attempted for the challenge, not a class). The function srl is not written necessarily in a practical manner, but within the constraints required by the problem.



      Questions



      Is there is a clearer, more straight-forward way to write this within the required constraints of the problem (perhaps with fewer ~ operations)?



      There is a need to avoid undefined behavior of the left shift, when k = 0. In this case int_bits - k = int_bits, which causes the shift to work unpredictably. Is there a better way to handle the undefined behavior of the shift operations, when the shift is larger than the number of bits in the interger?



      It seems to work correctly, but I lack an answer, so any feedback on the solution would be appreciated.



      Requirements





      • No additional right shifts or type casts may be used beyond the given expression



          /*Perform shift arithmetically*/
        unsigned xsra = (int) x >> k;


      • Only addition or subtraction may be used, no multiplication, division, or modulus


      • No comparison operators and no conditionals


      • Only bit-wise operations (except further right shifts) and logical operators may be used



      Code



      unsigned srl(unsigned x, int k) {
      /*Perform shift arithmetically*/
      unsigned xsra = (int) x >> k;

      unsigned int_bits = sizeof(int) << 3;//calculates the number of bits in int (assumes 8-bit byte)

      unsigned zero_or_all_bits = ~0 + !k;//for k = 0, corrects for the undefined behavior in
      //the left shift produced from int_bits - k = int_bits
      //if k != 0, !k == 0, and zero_or_all_bits == ~0
      //if k == 0, zero_or_all_bits == 0

      unsigned high_bit_mask = ~(zero_or_all_bits << (zero_or_all_bits & (int_bits - k)));
      /******************************************/
      //creates a mask of either all bits set in an unsigned int (UINT_MAX)
      //or a mask with k high bits cleared.
      //if k == 0, then high_bit_mask = ~(0 << 0) = ~0.
      //if k != 0, then high_bit_mask = ~(~0 << (~0 & (int_bits - k)))
      //ex. k == 3, high_bit_mask == 0x1FFFFFFF
      //ex. k == 0, high_bit_mask == 0xFFFFFFFF
      //ex. k == 31, high_bit_mask == 0xFFFFFFFE
      /******************************************/

      return xsra & high_bit_mask;


      }





      Test Code



      printf("Test srl:n");
      printf("srl(-1, 1): 0x%.8xn", srl(-1, 1));
      printf("srl(-1, 4): 0x%.8xn", srl(-1, 4));
      printf("srl(-1, 5): 0x%.8xn", srl(-1, 5));
      printf("srl(-1, 31): 0x%.8xn", srl(-1, 31));
      printf("srl(-1, 0): 0x%.8xn", srl(-1, 0));

      printf("srl(0x7FFFFFFF, 1): 0x%.8xn", srl(0x7FFFFFFF, 1));
      printf("srl(0x7FFFFFFF, 4): 0x%.8xn", srl(0x7FFFFFFF, 4));
      printf("srl(0x7FFFFFFF, 5): 0x%.8xn", srl(0x7FFFFFFF, 5));
      printf("srl(0x7FFFFFFF, 31): 0x%.8xn", srl(0x7FFFFFFF, 31));
      printf("srl(0x7FFFFFFF, 0): 0x%.8xn", srl(0x7FFFFFFF, 0));

      printf("srl(0x80000000, 1): 0x%.8xn", srl(0x80000000, 1));
      printf("srl(0x80000000, 4): 0x%.8xn", srl(0x80000000, 4));
      printf("srl(0x80000000, 5): 0x%.8xn", srl(0x80000000, 5));
      printf("srl(0x80000000, 31): 0x%.8xn", srl(0x80000000, 31));
      printf("srl(0x80000000, 0): 0x%.8xn", srl(0x80000000, 0));

      printf("srl(0, 1): 0x%.8xn", srl(0, 1));
      printf("srl(0, 4): 0x%.8xn", srl(0, 4));
      printf("srl(0, 5): 0x%.8xn", srl(0, 5));
      printf("srl(0, 31): 0x%.8xn", srl(0, 31));
      printf("srl(0, 0): 0x%.8xn", srl(0, 0));

      printf("srl(1, 1): 0x%.8xn", srl(1, 1));
      printf("srl(1, 4): 0x%.8xn", srl(1, 4));
      printf("srl(1, 5): 0x%.8xn", srl(1, 5));
      printf("srl(1, 31): 0x%.8xn", srl(1, 31));
      printf("srl(1, 0): 0x%.8xn", srl(1, 0));




      Output




      Test srl:

      srl(-1, 1): 0x7fffffff

      srl(-1, 4): 0x0fffffff

      srl(-1, 5): 0x07ffffff

      srl(-1, 31): 0x00000001

      srl(-1, 0): 0xffffffff

      srl(0x7FFFFFFF, 1): 0x3fffffff

      srl(0x7FFFFFFF, 4): 0x07ffffff

      srl(0x7FFFFFFF, 5): 0x03ffffff

      srl(0x7FFFFFFF, 31): 0x00000000

      srl(0x7FFFFFFF, 0): 0x7fffffff

      srl(0x80000000, 1): 0x40000000

      srl(0x80000000, 4): 0x08000000

      srl(0x80000000, 5): 0x04000000

      srl(0x80000000, 31): 0x00000001

      srl(0x80000000, 0): 0x80000000

      srl(0, 1): 0x00000000

      srl(0, 4): 0x00000000

      srl(0, 5): 0x00000000

      srl(0, 31): 0x00000000

      srl(0, 0): 0x00000000

      srl(1, 1): 0x00000000

      srl(1, 4): 0x00000000

      srl(1, 5): 0x00000000

      srl(1, 31): 0x00000000

      srl(1, 0): 0x00000001








      c integer bitwise






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      RJMRJM

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