Sum of digits C program to calculate the sum of digits of a number, we use modulus operator (%) to extract individual digits of a number and keep on adding them.
Sum of digits of a number in C
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 we don't recommend it.
Output of program:
For example, if the input is 98, the variable sum is 0 initially
98%10 = 8 (% is modulus operator, which gives us the 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 another one, 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.
C program to find sum of digits using for loop
int main()
{
int n, sum = 0, r;
printf("Enter a number\n");
for (scanf("%d", &n); n != 0; n = n/10) {
r = n % 10;
sum = sum + r;
}
printf("Sum of digits of a number = %d\n", sum);
return 0;
}
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 its every character into an integer and add all these integers.
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 huge, which we can't store in an int or a long long data type variable. See the example below.
Output of program:
9876543210
Sum of digits of 9876543210 = 45
C program to find sum of digits of a number using recursion
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) {
if (n == 0) // Base case
return 0;
return (n%10 + add_digits(n/10));
}