How to write a recursive method to return the sum of digits in an int?

Question

So this is my code so far.

    public int getsum (int n){
        int num = 23456;
        int total = 0;
        while (num != 0) {
            total += num         


        

Answer 1

I think it's the shortest so far. The input thing is up too you, though.

 public static int getSum(int input)  {  //example: input=246
       int sum=0;   
       if (input%10==input)  { //246%10=6;  
              return input%10; //2%10=2
       }

       return input%10+getSum((input-input%10)/10); //(246-6)/10=24; 24%10=4
 }

Answer 2

Try this:

int getSum(int num)
{
    total = total + num % 10;
    num = num/10;
    if(num == 0)
    {
        return total;
    } else {
        return getSum(num);
    }
}

Answer 3

int getSum(int N)
{
    int totalN = 0;

    totalN += (N% 10);
    N/= 10;

    if(N == 0)
        return totalN;
    else 
        return getSum(N) + totalN;
}

Answer 4

Here it is,

//sumDigits function
int sumDigits(int n, int sum) {    
    // Basic Case to stop the recursion
if (n== 0)  {
        return sum;
    } else {
        sum = sum + n % 10;  //recursive variable to keep the digits sum
        n= n/10;
        return sumDigits(n, sum); //returning sum to print it.
    }
}

An example of the function in action:

public static void main(String[] args) {
     int sum = sumDigits(121212, 0);
     System.out.println(sum);
}

Answer 5

#include <iostream>
int useRecursion(int x);
using namespace std;

int main(){
    int n;
    cout<<"enter an integer: ";
    cin>>n;
    cout<<useRecursion(n)<<endl;
    return 0;
}

int useRecursion(int x){
    if(x/10 == 0)
        return x;
    else
        return useRecursion(x/10) + useRecursion(x%10);
}

Answer 6

public static int digitSum (int n)
  { 
    int r = n%10;       //remainder, last digit of the number
    int num = n/10;     //the rest of the number without the last digit
    if(num == 0)
    {
      return n;
    } else {
      return digitSum (num) + r;
    }}