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  4. Not able to find the square of the floating number ...

Not able to find the square of the floating number ...

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  • C CPallini

    Quote:

    double sqrt (double f) { double x, z, tempf; unsigned long *tfptr = ((unsigned long *)&tempf) + 1; tempf = f; *tfptr = (0xbfcdd90a - *tfptr)>>1; x = tempf; z = f*0.5; x = (1.5*x) - (x*x)*(x*z); //The more you make replicates of this statement //the higher the accuracy, here only 2 replicates are used x = (1.5*x) - (x*x)*(x*z); return x*f; }

    That is really messy! You are using uninitialized variables (that is garbage). It looks you want to implement the Babylonian method[^] but you do nothing for computing the initial guess. The following code is based on that very Wikipedia page.

    void initial_guess(double r, int *pa, int *pn)
    {
    *pn = 0;

    while ( r < 1.0)
    {
    r *= 100.0;
    --(*pn);
    }
    while ( r >= 100.0 )
    {
    r /= 100.0;
    ++(*pn);
    }

    *pa = (r < 10.0) ? 2 : 6;
    }

    double square_root(double r)
    {
    int a, i, n;
    initial_guess(r, &a, &n);
    double x=1.0;

    while (n < 0)
    {
    x/=10.0;
    ++n;
    }
    while(n > 0)
    {
    x *= 10.0;
    --n;
    }

    x *= a;

    for (i = 0; i<10; ++i) // 10 is the arbitrary number of iterations I chose
    {
    x = 0.5 * (x + r/x);
    }

    return x;
    }

    THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

    M Offline
    M Offline
    mybm1
    wrote on last edited by
    #6

    thanks....

    C 1 Reply Last reply
    0
    • M mybm1

      thanks....

      C Offline
      C Offline
      CPallini
      wrote on last edited by
      #7

      You are welcome.

      THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

      1 Reply Last reply
      0
      • C CPallini

        Quote:

        double sqrt (double f) { double x, z, tempf; unsigned long *tfptr = ((unsigned long *)&tempf) + 1; tempf = f; *tfptr = (0xbfcdd90a - *tfptr)>>1; x = tempf; z = f*0.5; x = (1.5*x) - (x*x)*(x*z); //The more you make replicates of this statement //the higher the accuracy, here only 2 replicates are used x = (1.5*x) - (x*x)*(x*z); return x*f; }

        That is really messy! You are using uninitialized variables (that is garbage). It looks you want to implement the Babylonian method[^] but you do nothing for computing the initial guess. The following code is based on that very Wikipedia page.

        void initial_guess(double r, int *pa, int *pn)
        {
        *pn = 0;

        while ( r < 1.0)
        {
        r *= 100.0;
        --(*pn);
        }
        while ( r >= 100.0 )
        {
        r /= 100.0;
        ++(*pn);
        }

        *pa = (r < 10.0) ? 2 : 6;
        }

        double square_root(double r)
        {
        int a, i, n;
        initial_guess(r, &a, &n);
        double x=1.0;

        while (n < 0)
        {
        x/=10.0;
        ++n;
        }
        while(n > 0)
        {
        x *= 10.0;
        --n;
        }

        x *= a;

        for (i = 0; i<10; ++i) // 10 is the arbitrary number of iterations I chose
        {
        x = 0.5 * (x + r/x);
        }

        return x;
        }

        THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

        M Offline
        M Offline
        mybm1
        wrote on last edited by
        #8

        using

        for(i=0;i<5;i++)
        { //float y,sum;
        float f= ((float)rand()/(float)(RAND_MAX))*a;
        y=square(f);
        sum +=y;
        }
        printf("\n summation of squared number is:=%f\t",sum);

        gives output but not the exact answer.. any solution actually i want to find the summation of that squared float number..

        C 1 Reply Last reply
        0
        • M mybm1

          using

          for(i=0;i<5;i++)
          { //float y,sum;
          float f= ((float)rand()/(float)(RAND_MAX))*a;
          y=square(f);
          sum +=y;
          }
          printf("\n summation of squared number is:=%f\t",sum);

          gives output but not the exact answer.. any solution actually i want to find the summation of that squared float number..

          C Offline
          C Offline
          CPallini
          wrote on last edited by
          #9

          How do you check the result?

          THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

          M 1 Reply Last reply
          0
          • C CPallini

            How do you check the result?

            THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

            M Offline
            M Offline
            mybm1
            wrote on last edited by
            #10

            Quote:

            Generated random number || Squaring of random number: ========================================================= 0.035679 ----- 0.001273 0.079935 ----- 0.006390 0.033320 ----- 0.001110 0.042424 ----- 0.001800 0.012387 ----- 0.000153 summation of squared number is:=0.010346

            where the actual summation should be 0.10726

            C J 2 Replies Last reply
            0
            • M mybm1

              Quote:

              Generated random number || Squaring of random number: ========================================================= 0.035679 ----- 0.001273 0.079935 ----- 0.006390 0.033320 ----- 0.001110 0.042424 ----- 0.001800 0.012387 ----- 0.000153 summation of squared number is:=0.010346

              where the actual summation should be 0.10726

              C Offline
              C Offline
              CPallini
              wrote on last edited by
              #11

              You are wrong, the output of the following program:

              int main()
              {
              int i;
              double sum = 0.0;
              double a[] =
              {
              0.035679,
              0.079935,
              0.03332,
              0.042424,
              0.012387
              };

              for (i=0; i<sizeof(a)/sizeof(a[0]); ++i)
              {
              double sq = square_root(a[i]);
              printf("a[%d] = %g, sqrt(a[%d])=%g\n", i, a[i], i, sq);
              sum += sq;
              }
              printf("sum of square roots = %g\n", sum);
              }

              is

              a[0] = 0.035679, sqrt(a[0])=0.188889
              a[1] = 0.079935, sqrt(a[1])=0.282728
              a[2] = 0.03332, sqrt(a[2])=0.182538
              a[3] = 0.042424, sqrt(a[3])=0.205971
              a[4] = 0.012387, sqrt(a[4])=0.111297
              sum of square roots = 0.971422

              That is correct (at least according to Excel :-) ).

              THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

              M 1 Reply Last reply
              0
              • M mybm1

                Quote:

                Generated random number || Squaring of random number: ========================================================= 0.035679 ----- 0.001273 0.079935 ----- 0.006390 0.033320 ----- 0.001110 0.042424 ----- 0.001800 0.012387 ----- 0.000153 summation of squared number is:=0.010346

                where the actual summation should be 0.10726

                J Offline
                J Offline
                jeron1
                wrote on last edited by
                #12

                Are you needing the sqaure or the square root? PS and the sum of your numbers is 0.010726.

                "the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst

                M 1 Reply Last reply
                0
                • C CPallini

                  You are wrong, the output of the following program:

                  int main()
                  {
                  int i;
                  double sum = 0.0;
                  double a[] =
                  {
                  0.035679,
                  0.079935,
                  0.03332,
                  0.042424,
                  0.012387
                  };

                  for (i=0; i<sizeof(a)/sizeof(a[0]); ++i)
                  {
                  double sq = square_root(a[i]);
                  printf("a[%d] = %g, sqrt(a[%d])=%g\n", i, a[i], i, sq);
                  sum += sq;
                  }
                  printf("sum of square roots = %g\n", sum);
                  }

                  is

                  a[0] = 0.035679, sqrt(a[0])=0.188889
                  a[1] = 0.079935, sqrt(a[1])=0.282728
                  a[2] = 0.03332, sqrt(a[2])=0.182538
                  a[3] = 0.042424, sqrt(a[3])=0.205971
                  a[4] = 0.012387, sqrt(a[4])=0.111297
                  sum of square roots = 0.971422

                  That is correct (at least according to Excel :-) ).

                  THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

                  M Offline
                  M Offline
                  mybm1
                  wrote on last edited by
                  #13

                  this is different program its squaring of number not square root...

                  C 1 Reply Last reply
                  0
                  • J jeron1

                    Are you needing the sqaure or the square root? PS and the sum of your numbers is 0.010726.

                    "the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst

                    M Offline
                    M Offline
                    mybm1
                    wrote on last edited by
                    #14

                    Actually i need square but its solve i got it but i am not able to get the summation value of the square number correctly .plz help me out..

                    1 Reply Last reply
                    0
                    • M mybm1

                      this is different program its squaring of number not square root...

                      C Offline
                      C Offline
                      CPallini
                      wrote on last edited by
                      #15

                      How do you compute the squares? This program:

                      #include <stdio.h>
                      int main()
                      {
                      double a [] =
                      {
                      0.035679,
                      0.079935,
                      0.033320,
                      0.042424,
                      0.012387
                      };
                      double sum, square;
                      int n;

                      sum = 0.0;
                      for (n=0; n<sizeof(a)/sizeof(a[0]); ++n)
                      {
                      square = a[n]*a[n];
                      sum += square;
                      printf("%g ----> %g\n", a[n], square);
                      }

                      printf("sum of squares: %g\n", sum);

                      return 0;
                      }

                      gives:

                      0.035679 ----> 0.00127299
                      0.079935 ----> 0.0063896
                      0.03332 ----> 0.00111022
                      0.042424 ----> 0.0017998
                      0.012387 ----> 0.000153438
                      sum of squares: 0.0107261

                      THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

                      1 Reply Last reply
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