# How do I write a recursive method to return the sum of digits to an int?

So this is my code.

``````    public int getsum (int n){
int num = 23456;
int total = 0;
while (num != 0) {
total += num % 10;
num /= 10;
}
}
```

```

The problem is I cant / know how to change this to a recursive method Im kind of new with recursion and I need help implementing this method to change it so that it is recursive.

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12 replies

Short, recursive and does the job:

``````int getsum(int n) {
return n == 0 ? 0 : n % 10 + getsum(n/10);
}
```

```
+13

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Here he 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 a function in action:

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

```
+6

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``````public int sumDigits(int n) {
return (n - 1) % 9 + 1;
}
```

```
+2

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``````public static int sumOfDigit(int num){
int sum=0;
if (num == 0)
return sum;
sum = num%10 + sumOfDigit(num/10);
return sum;
}
public static void main(String args[]) {
Scanner input=new Scanner(System.in);
System.out.print("Input num : ");
int num=input.nextInt();
int s=sumOfDigit(num);
System.out.println("Sum = "+s);
}
```

```

}

+1

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Try the following:

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

```
0

source

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

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

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

```
0

source

``````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;
}}
```

```
0

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This works for positive numbers.

``````public int sumDigits(int n) {
int sum = 0;
if(n == 0){
return 0;
}
sum += n % 10; //add the sum
n /= 10; //keep cutting
return sum + sumDigits(n); //append sum to recursive call
}
```

```
0

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``````    import java.util.Scanner;
public static void main(String[] args){
Scanner input = new Scanner(System.in);
System.out.print("Enter a number: ");
System.out.println();
int number = input.nextInt();
System.out.println("The sum of the digits is " +adder(number));

}
int length = String.valueOf(num).length();
int first , last , sum;
if (length==1){
return num;
}
else
{
first = num /10;
last = num % 10;
}
return sum;
}
}
```

```
0

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I see many solutions here, but none seem to be as simple as what follows. I've tested it countless times and this is not a problem:

``````public int sumDigits(int n) {
if (n == 0){
return 0;
}
else{
return n%10 + sumDigits(n/10);
}
}
```

```
0

source

``````#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);
}
```

```
-1

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I think this is the shortest option. However, the input thing is you too.

`````` 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
}
```

```
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