Thread: averaging

I am trying to calculate the average of the numbers the user entered but I can't figure out how many numbers the user actually entered to so I can divide by the sum. The user can enter up to 10 numbers or a zero to exit at anytime. I need to be able to calculate the average but not include zero.

Code:

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

int main(void)
{
  char theNums[10][50];
  int nums[10];
  int i, j, sum = 0;
  int average, qty;

  printf("Start by entering up to 10 numbers.  To exit enter a zero.\n");
  printf("\n");
	
  for(i=0; i < 10; i++)
  {
    printf("Enter a Number: ");
    fgets(theNums[i], 50, stdin);
		
    j = strlen(theNums[i])-1;
		
    theNums[i][j] = '\0'; /* get rid of '\n' */
		
    if(theNums[i][0] == '0')
      break;
  else if(!isdigit(theNums[i][0]))
  {
    printf("%s is not a valid entry.\n", theNums[i]);
    nums[i] = 0;
  }
  else		
    nums[i] = atoi(theNums[i]);
  }
	
  printf("\n\n");
	
  for(j=0; j < i; j++)
    sum+=nums[j];
    average = (sum/qty);

  printf("The sum of your numbers is %d", sum);
  printf("\n");

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

  getchar(); /*wait for keystroke before exiting*/
	
  return 0;
}