Thread: Recursion to find sum of digits of n digit number ?

Hi,
The See the below recursion.

Code:

#include<stdio.h>
int rec(int);
int main()
{
    int n, b; 
    scanf("%d", &n);
    b = rec (n);
    printf("%d", b);    
    return 0;
}
int rec(int n)
{
    
    if (n==0)
    return 0;
    else
    return (n%10+rec(n/10));
}

it works, but:

Code:

#include<stdio.h>
void rec(int);
int main()
{
    int n;     
    scanf("%d", &n);
    rec (n);    
    return 0;
}
void rec(int n)
{    
    if (n==0)
    rec(n)=0;
    else
    printf("%d", (n%10+rec(n/10)));
}

This is not working, is it possible to make it work?

thanks.

Code:

void rec(int);

If the return type of a function is void it cannot return any value. These instructions are invalid:

Originally Posted by justine

Code:

rec(n)=0;
printf("%d", (n%10+rec(n/10)));

Originally Posted by justine

This is not working, is it possible to make it work?

Yes. You can define a static variable which will hold the sum of digits.

Code:

#include<stdio.h>
int rec(int);
int main()
{
    int n;    
    scanf("%d", &n);
    rec (n);   
    return 0;
}
int rec(int n)
{   
    static int k;
    if (n==0)
    return 0;
    else
    {
        k=k+n%10;
        rec(n/10);
    }    
    printf("%d\n", k);
}

It prints many time the result.

I'm confused with what you want to do. You have a working recursive program, but you don't want to use it. Do you want a non-recursive version of it (ie iterative version)?

The first recursive version is as short and as clear as it can be. I guarantee you, you can't improve it. What's the goal here?

@DRK I couldn't follow you, can you please give full program. @Adak I was just trying to print the result in function.

Originally Posted by justine

@DRK I couldn't follow you, can you please give full program.

Eh, since you've almost done it yourself...

Code:

#include<stdio.h>

void rec(int n);

int main()
{
    int n;
    scanf("%d", &n);
    rec(n);
    return 0;
}

void rec(int n)
{
    static int k;
    if (n == 0)
    {
        printf("%d\n", k);
        k = 0;
    }
    else
    {
        k += n % 10;
        rec(n / 10);
    }
}

Thanks DRK