It is supposed to take the number of subjects as input and take marks according to it and print the average of the grades , But it is not working....n

Question

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

int main()
{
   int n, tg=0;
   printf("Enter the number of subjects : ");
   scanf("%d", &n);

   while (n>0)
   {
      int grade;
      printf("Enter the marks : ");
      scanf("%d", &grade);
      n--;
      tg+=grade;
   }
   float avg;
   avg = (tg/n);
   printf("\nYour Average Grade is %f\n", avg);
   return 0;
}

/* It is supposed to show the average grade but it is not displaying any result, but displays this instead "Process returned -1073741676 (0xC0000094) execution time: 6.532 s"

Your code reduces `n` to zero and then evaluates `tg/n`, so the program crashes. — Eric Postpischil, Dec 19 '21 at 08:25
Also, `tg/n` uses `int` arithmetic. To perform floating-point division, one of the operands must be floating-point. — Eric Postpischil, Dec 19 '21 at 08:26
Does this answer your question? [What is the behavior of integer division?](https://stackoverflow.com/questions/3602827/what-is-the-behavior-of-integer-division) — phuclv, Dec 19 '21 at 08:43

Vlad from Moscow · Answer 1 · 2021-12-19T08:42:53.450

0

After the while loop

while (n>0)
{
   //...
   n--;
}

the variable n is equal tp 0. So this statement

 avg = (tg/n);

throws exception dividing by zero.

You could change the while loop for a for loop like for example

int n = 0;

//...

for ( int i = 0; i < n; i++ )
//..

and then after the loop

 avg = n > 0 ? ( float )tg/n : 0.0f;

edited Dec 19 '21 at 08:42

answered Dec 19 '21 at 08:36

Vlad from Moscow

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It is supposed to take the number of subjects as input and take marks according to it and print the average of the grades , But it is not working....n

1 Answers1