Sum of digits of a Number in C

June 29, 2023

C Program to Find Sum of Digits of a Number 

In this C program, we will code Sum of Digits of a Number in C we will allow the user to enter any number and then we will divide the number into individual digits and add those individuals (sum=sum+digit) digits using While Loop.

Ex:-  number is 231456

 2 + 3 + 1 + 4 + 5 + 6 = 21

Sum of digit of a given number is 21

sum of digits of a number in c

Sum of Digits in C Program

Working:-

For user input num

  • Initialize sum = 0
  • Extract the last digit of the number and add it to the sum (sum += num % 10)
  •  Reduce the number  (num = num / 10
  • Keep doing this until num becomes 0
Sum Of Digits Of A Number in C

Method 1

Uses the iterative way of calculation

C Code

Run

// Write a Program to Find Sum of Digits of a given number
#include<stdio.h>

int main ()
{
    int num, sum = 0;
 
    num = 1234;
    printf("The number is = %d\n",num);

    //loop to find sum of digits
    while(num!=0){
        sum += num % 10;
        num = num / 10;
    }
 
    //output
    printf("Sum: %d\n",sum);
 
    return 0;

}
// Time complexity : O(N)
// Space complexity : O(1)

Output

The number is = 1234
Sum: 10

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While loop in C

Method 2

This method uses a recursive approach in C

Code

Run

// Time complexity : O(N)
// Space complexity : O(1)
// Auxiliary Space Complexity : O(N)
// Due to function call stack
#include<stdio.h>

int getSum(int num, int sum){
    
    if(num == 0)
        return sum;

    sum += num % 10;

    return getSum(num/10, sum);
}
int main ()
{
    int num, sum = 0;
    num = 6789;
    printf("The number is: %d\n",num);   
    printf("Sum: %d\n",getSum(num, sum));
 
    return 0;

}
// Time complexity : O(N)
// Space complexity : O(1)
// Auxilary space complexity O(N)

Output

The number is = 6789
Sum: 30
While loop in C

Method 3

The above recursive method can also be reduced down to something like this below -. Note this method also uses a recursive approach in C

Code

Run
// Time complexity : O(N)
// Space complexity : O(1)
// Auxiliary Space Complexity : O(N)
// Due to function call stack
#include<stdio.h>
 
int getSum(int num)
{
    if(num == 0){
        return 0 ;
    }
    
    int digit = num % 10;
    
    return digit + getSum(num / 10) ;
}
 
int main()
{
    int num = 2545;
    printf("Sum: %d", getSum(num));
    return 0;
}

Output

Sum: 16

Method 4

We can also take numbers as input in char array/string format.

The below program uses this approach. You must have knowledge of the Ascii Table.

Please check ASCII Table here

Code

Run

// Time Complexity : O(N)
// Space Complexity : O(1)
#include<stdio.h>
#include<string.h>

int getSum(char str[], int len)
{
    int sum = 0;
 
    // Traverse through the whole string(char array)
    for (int i = 0; i < len; i++) 
    {
        // Ascii value pf numbers start from 48
        // subtracting 48 will give us value in int
        sum = sum + str[i] - 48;
    }
    return sum;
}
 
int main()
{
    char num[] = "987654321";
    int len = strlen(num);
    
    // can also use below
    // int len = sizeof(num)/sizeof(num[0]);
    
    printf("Sum: %d", getSum(num, len));
    return 0;
}

Output

Sum: 45

Method 5

Another method working with character input. You must have knowledge of the Ascii Table.

Please check ASCII Table here

Code

Run

#include<stdio.h> 
 
int main()
{
    int index = 0, sum = 0, temp;
    char num[1000];
   
    // taking character input for num
    printf("Input an integer: ");
    scanf("%s", num);
 
    // continue until we hit end of the char array 
    while (num[index] != '\0') 
    {
        // subtract ascii value of num from ascii value of '0'
        temp   = num[index] - '0';
        sum = sum + temp;
        index++;
    }
 
   printf("Sum of digits of %s is: %d ",num, sum);
 
   return 0;
}

Output

Input an integer: 12345
Sum of digits of 12345 is: 15

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