what value does 'return 0;' statement exactly return in user defined functions in C++

Question

I was going through this recursion program in cpp on w3schools.

#include <iostream>
using namespace std;

int sum(int k) {
  if (k > 0) {
    return k + sum(k - 1);
  } else {
    return 0;
  }
}

int main() {
  int result = sum(10);
  cout << result;
  return 0;
}

There is a condition block in the recursive function above. Assuming that it will eventually go to the 'else' block, the value of 0 will be returned from the function. However, it's confusing to me how the function call sum(10) is working properly and not just returning 0.

Could someone explain the recursion to me.

Answer 1

Expanding on my comments, there's nothing special with a recursive function call. It's just like any other function call.

Lets take the following example:

int c(int k)
{
    return 0;
}

int b(int k)
{
    return k + c(k - 1);
}

int a(int k)
{
    return k + b(k - 1);
}

int main(void)
{
    std::cout << a(2) << '\n';
}

The call a(2) will do return 2 + b(1);
The call b(1) will do return 1 + c(0);
The call c(0) will do return 0;

This is really the same as the call sum(2) from the recursive example in the question:

The call sum(2) will do return 2 + sum(1);
The call sum(1) will do return 1 + sum(0);
The call sum(0) will do return 0;

The result is 2 + 1 + 0 which equals 3.