Sum of digits

#include <stdio.h>

int main()
{
   int n, t, sum = 0, remainder;

   printf("Enter an integer\n");
   scanf("%d", &n);

   t = n;

   while (t != 0)
   {
      remainder = t % 10;
      sum       = sum + remainder;
      t         = t / 10;
   }

   printf("Sum of digits of %d = %d\n", n, sum);

   return 0;
}

If you wish you can modify the input variable (n) and without using an additional variable (t) but it isn't recommended.

Output of program:

Calculate sum of digits in C without modulus operator

C program to find the sum of digit(s) of an integer that does not use modulus operator. Our program uses a character array (string) for storing an integer. We convert every character of the string into an integer and add all these integers.

#include <stdio.h>

int main()
{
   int c, sum, t;
   char n[1000];
 
   printf("Input an integer\n");
   scanf("%s", n);

   sum = c = 0;
 
   while (n[c] != '\0') {
      t   = n[c] - '0'; // Converting character to integer
      sum = sum + t;
      c++;
   }

   printf("Sum of digits of %s = %d\n", n, sum);

   return 0;
}

An advantage of this method is that the input integer can be very large which can't be stored in an int or a long long variable, see an example below.

Output of program:

Input an integer
123456789123456789123456789
Sum of digits of 123456789123456789123456789 = 135

Sum of digits of a number C program using recursion

#include <stdio.h>

int add_digits(int);

int main()
{
  int n, result;

  scanf("%d", &n);

  result = add_digits(n);

  printf("%d\n", result);

  return 0;
}

int add_digits(int n) {
  static int sum = 0;

  if (n == 0) {
    return 0;
  }

  sum = n%10 + add_digits(n/10);

  return sum;
}

The static variable sum is used and initialized to 0; its value will persist after function calls, i.e., it is initialized once when the function is called for the first time.