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Home/ Questions/Q 8713993
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T05:28:55+00:00 2026-06-13T05:28:55+00:00

If I leave all variables as int I get 32. The division is supposed

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If I leave all variables as int I get 32. The division is supposed to give me 32.5 so I thought that changing everything to double would do it, but it just gives me zero…

Here is the code (everything in int):

#include <stdio.h>
#include <stdlib.h>

void sommeTableau(int tableau[], int tailleTableau, int *x, int *average);

int main(int argc, char *argv[])
{
    int tableau[4] = {30, 50, 50};
    int *x = 0;
    int *average = 0;
    const int tailleTab = 4;

    sommeTableau(tableau, tailleTab, &x, &average);

    printf("The average is %d\n", average);

    return 0;
}

void sommeTableau(int tableau[], int tailleTableau, int *x, int *average)
{
  int i = 0;

  for (i = 0 ; i < tailleTableau ; i++)
  {
  *x = *x + tableau[i];
  }
  *average = *x/tailleTableau;

}

So this works and gives me 32… Now if I change everything to double and %d to %f in the printf function, it gives me zero and I don’t understand why…

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T05:28:56+00:00Added an answer on June 13, 2026 at 5:28 am

    There is lots of mistakes in your code. I corrected all of them

    #include <stdio.h>
    #include <stdlib.h>
    
    void sommeTableau(int tableau[], int tailleTableau, int *x, double *average);
    
    int main(int argc, char *argv[])
    {
        int tableau[4] = {30, 50, 50};
        int x = 0;
        double average = 0;
        const int tailleTab = 4;
    
        sommeTableau(tableau, tailleTab, &x, &average);
    
        printf("The average is %lf\n", average);
    
        return 0;
    }
    
    void sommeTableau(int tableau[], int tailleTableau, int *x, double *average)
    {
      int i = 0;
    
      for (i = 0 ; i < tailleTableau ; i++)
      {
      *x = *x + tableau[i];
      }
      *average = (double)(*x)/(double)tailleTableau;
    
    }
    
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