What is the output of this code & most importantly HOW?

Question

int foo(int n, int r){
if(n>0)
return (n % r + foo(n / r, r));
else
return 0;
}
int main(){
printf("%d", foo(345,10));
}

Answer 1

foo(345, 10)

5 + foo(34, 10)

5 + 4 + foo(3, 10)

5 + 4 + 3 + foo(0, 10)

5 + 4 + 3 + 0

= 12

edit:
first step is included.

Answer 2

Ashfaq Reshad

It is a code to calculate sum of 345

After function calling first it check the condition if(n>0) in which
recursive returns 3 times as -
345/10 = 34 /10= 3/10=3
3%10=3
34%10=4
345%10=5
3+4+5=12

Answer 3

This program shows a recursive function foo, which will call itself until n has become less than Zero...

In the function it adds the modulo of n and r to the result of the next call...
Now lets Go through it:
1. call: 345 % 10 = 5
2. call: 34 % 10 = 4
3. call: 3 % 10 = 3
4. call: 0

Add together: 12 (which ist the output of the code)..

Note, that n is shortend by one digit from right in every next call by integer Division by 10...