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Sum of digits in C

C program to calculate the sum of digits of a number: we will use modulus operator (%) to extract individual digits of a number and keep on adding them.

C programming code

#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:
Sum of digits C program output

For example, if the input is 98, the variable sum is 0 initially
98%10 = 8 (% is modulus operator which gives us remainder when 98 is divided by 10).
sum = sum + remainder
so sum = 8 now.
98/10 = 9 because in C language whenever we divide an integer by an another integer we get an integer.
9%10 = 9
sum = 8 (previous value) + 9
sum = 17
9/10 = 0.
So finally n = 0, the loop ends we get the required sum.

Download Add digits program.

Calculate sum of digits in C without modulus operator

C program to find the sum of digit(s) of an integer which 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 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.