Program for Sum of the digits of a given number

Difficulty Level : Easy
Last Updated : 04 Mar, 2021

Given a number, find sum of its digits.

Examples :

Input : n = 687
Output : 21

Input : n = 12
Output : 3

General Algorithm for sum of digits in a given number:

Get the number
Declare a variable to store the sum and set it to 0
Repeat the next two steps till the number is not 0
Get the rightmost digit of the number with help of remainder ‘%’ operator by dividing it with 10 and add it to sum.
Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
Print or return the sum

Below are the solutions to get sum of the digits.
1. Iterative:

C++

// C program to compute sum of digits in
// number.
#include <iostream>
using namespace std;
 
/* Function to get sum of digits */
class gfg {
public:
    int getSum(int n)
    {
        int sum = 0;
        while (n != 0) {
            sum = sum + n % 10;
            n = n / 10;
        }
        return sum;
    }
};
 
// Driver code
int main()
{
    gfg g;
    int n = 687;
    cout << g.getSum(n);
    return 0;
}
// This code is contributed by Soumik

C

// C program to compute sum of digits in
// number.
#include <stdio.h>
 
/* Function to get sum of digits */
int getSum(int n)
{
    int sum = 0;
    while (n != 0) {
        sum = sum + n % 10;
        n = n / 10;
    }
    return sum;
}
 
// Driver code
int main()
{
    int n = 687;
    printf(" %d ", getSum(n));
    return 0;
}

Java

// Java program to compute
// sum of digits in number.
import java.io.*;
 
class GFG {
 
    /* Function to get sum of digits */
    static int getSum(int n)
    {
        int sum = 0;
 
        while (n != 0) {
            sum = sum + n % 10;
            n = n / 10;
        }
 
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 687;
 
        System.out.println(getSum(n));
    }
}
 
// This code is contributed by Gitanjali

Python3

# Python 3 program to
# compute sum of digits in
# number.
 
# Function to get sum of digits
 
 
def getSum(n):
 
    sum = 0
    while (n != 0):
 
        sum = sum + int(n % 10)
        n = int(n/10)
 
    return sum
 
 
# Driver code
n = 687
print(getSum(n))

C#

// C# program to compute
// sum of digits in number.
using System;
 
class GFG {
    /* Function to get sum of digits */
    static int getSum(int n)
    {
        int sum = 0;
 
        while (n != 0) {
            sum = sum + n % 10;
            n = n / 10;
        }
 
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 687;
        Console.Write(getSum(n));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP Code to compute sum 
// of digits in number.
 
// Function to get 
// $sum of digits 
function getsum($n)
{ 
    $sum = 0;
    while ($n != 0)
    {
        $sum = $sum + $n % 10;
        $n = $n/10;
    }
    return $sum;
}
 
// Driver Code
$n = 687;
$res = getsum($n);
echo("$res");
 
// This code is contributed by 
// Smitha Dinesh Semwal.
?>

Output

How to compute in a single line?
Below function has three lines instead of one line but it calculates sum in line. It can be made one line function if we pass pointer to sum.

C++

#include <iostream>
using namespace std;
 
/* Function to get sum of digits */
class gfg {
public:
    int getSum(int n)
    {
        int sum;
 
        /* Single line that calculates sum */
        for (sum = 0; n > 0; sum += n % 10, n /= 10)
            ;
 
        return sum;
    }
};
 
// Driver code
int main()
{
    gfg g;
    int n = 687;
    cout << g.getSum(n);
    return 0;
}
// This code is contributed by Soumik

C

#include <stdio.h>
 
/* Function to get sum of digits */
int getSum(int n)
{
    int sum;
 
    /* Single line that calculates sum */
    for (sum = 0; n > 0; sum += n % 10, n /= 10)
        ;
 
    return sum;
}
 
// Driver code
int main()
{
    int n = 687;
    printf(" %d ", getSum(n));
    return 0;
}

Java

// Java program to compute
// sum of digits in number.
import java.io.*;
 
class GFG {
 
    /* Function to get sum of digits */
    static int getSum(int n)
    {
        int sum;
 
        /* Single line that calculates sum */
        for (sum = 0; n > 0; sum += n % 10, n /= 10)
            ;
 
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 687;
 
        System.out.println(getSum(n));
    }
}
 
// This code is contributed by Gitanjali

Python3

# Function to get sum of digits
 
 
def getSum(n):
 
    sum = 0
 
    # Single line that calculates sum
    while(n > 0):
        sum += int(n % 10)
        n = int(n/10)
 
    return sum
 
 
# Driver code
n = 687
print(getSum(n))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to compute
// sum of digits in number.
using System;
 
class GFG {
    static int getSum(int n)
    {
        int sum;
 
        /* Single line that calculates sum */
        for (sum = 0; n > 0; sum += n % 10, n /= 10)
            ;
 
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 687;
        Console.Write(getSum(n));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP Code for Sum the 
// digits of a given number
 
// Function to get sum of digits 
function getsum($n)
{
 
    // Single line that calculates $sum
    for ($sum = 0; $n > 0; $sum += $n % 10, 
                                  $n /= 10);
    return $sum;
}
 
// Driver Code
$n = 687;
echo(getsum($n));
 
// This code is contributed by 
// Smitha Dinesh Semwal.
?>

Output

2. Recursive
Thanks to ayesha for providing the below recursive solution.

C++

// C++ program to compute
// sum of digits in number.
#include <iostream>
using namespace std;
class gfg {
public:
    int sumDigits(int no)
    {
        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
    }
};
 
// Driver code
int main(void)
{
    gfg g;
    cout << g.sumDigits(687);
    return 0;
}

C

// C program to compute
// sum of digits in number.
#include <stdio.h>
using namespace std;
int sumDigits(int no)
{
    return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
}
 
int main(void)
{
    printf("%d", sumDigits(687));
    return 0;
}

Java

// Java program to compute
// sum of digits in number.
import java.io.*;
 
class GFG {
 
    /* Function to get sum of digits */
    static int sumDigits(int no)
    {
        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(sumDigits(687));
    }
}
 
// This code is contributed by Gitanjali

Python3

# Python program to compute
# sum of digits in number.
 
 
def sumDigits(no):
    return 0 if no == 0 else int(no % 10) + sumDigits(int(no/10))
 
 
# Driver code
print(sumDigits(687))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to compute
// sum of digits in number.
using System;
 
class GFG {
    /* Function to get sum of digits */
    static int sumDigits(int no)
    {
        return no == 0 ? 0 : no % 10 + sumDigits(no / 10);
    }
 
    // Driver code
    public static void Main()
    {
        Console.Write(sumDigits(687));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP program to compute 
// sum of digits in number.
function sumDigits($no)
{
return $no == 0 ? 0 : $no % 10 + 
                      sumDigits($no / 10) ;
}
 
// Driver Code
echo sumDigits(687);
 
// This code is contributed by aj_36
?>

Output

3.Taking input as String

When the number of digits of that number exceeds 10¹⁹ , we can’t take that number as integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character(in this case it is numeric)

Below is the implementation of the above approach

C++14

// C++ implementation of the above aprroach
#include <iostream>
using namespace std;
int getSum(string str)
{
    int sum = 0;
 
    // Traversing through the string
    for (int i = 0; i < str.length(); i++) {
        // Since ascii value of
        // numbers starts from 48
        // so we subtract it from sum
        sum = sum + str[i] - 48;
    }
    return sum;
}
 
// Driver Code
int main()
{
    string st = "123456789123456789123422";
    cout << getSum(st);
    return 0;
}

Java

// Java implementation of the above aprroach
import java.io.*;
class GFG {
 
    static int getSum(String str)
    {
        int sum = 0;
 
        // Traversing through the string
        for (int i = 0; i < str.length(); i++) {
 
            // Since ascii value of
            // numbers starts from 48
            // so we subtract it from sum
            sum = sum + str.charAt(i) - 48;
        }
        return sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String st = "123456789123456789123422";
        System.out.print(getSum(st));
    }
}
 
// This code is contributed by Dharanendra L V.

Python3

# Python implementation of the above aprroach
def getSum(n):
    # Initializing sum to 0
    sum = 0
    # Traversing through string
    for i in n:
        # Converting char to int
        sum = sum + int(i)
 
    return sum
 
 
n = "123456789123456789123422"
print(getSum(n))

C#

// C# implementation of the above aprroach
using System;
public class GFG {
    static int getSum(String str)
    {
        int sum = 0;
 
        // Traversing through the string
        for (int i = 0; i < str.Length; i++) {
 
            // Since ascii value of
            // numbers starts from 48
            // so we subtract it from sum
            sum = sum + str[i] - 48;
        }
        return sum;
    }
 
    // Driver Code
    static public void Main()
    {
        String st = "123456789123456789123422";
        Console.Write(getSum(st));
    }
}
 
// This code is contributed by Dharanendra L V.

Output

4. Using Tail Recursion

This problem can also be solved using Tail Recursion. Here is approach to solve it.

1. Add another variable “val” to the function and initialize it to ( val = 0 )

2. On every call to the function add the mod value (n%10) to the variable as “(n%10)+val” which is the last digit in n. Along with pass the variable n as n/10.

3. So on the First call it will have the last digit. As we are passing n/10 as n, It follows until n is reduced to single digit.

4. n<10 is the base case so When n < 10, then add the n to the variable as it is the last digit and return the val which will have the sum of digits

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class sum_of_digits {
   
    // Function to check sum
    // of digit using tail recursion
    static int sum_of_digit(int n, int val)
    {
        if (n < 10) {
            val = val + n;
            return val;
        }
        return sum_of_digit(n / 10, (n % 10) + val);
    }
 
    // Driven Program to check above
    public static void main(String args[])
    {
        int num = 12345;
        int result = sum_of_digit(num, 0);
        System.out.println("Sum of digits is " + result);
    }
}

Output

Sum of digits is 15

Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.

Attention reader! Don’t stop learning now. Get hold of all the important C++ Foundation and STL concepts with the C++ Foundation and STL courses at a student-friendly price and become industry ready.

My Personal Notes

Related Articles

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python3

C#

PHP

C++14

Java

Python3

C#

Java