Help me understand this recursion please

[cspctec]

I have the following code:

#include <iostream>

using namespace std;

int count_digits(int n)
{
    if (n >= 0 && n < 10)
    {
        return 1;
    }

    else if (n >= 10)
    {
        return 1 + count_digits(n / 10);
    }
}

int main()
{
    cout << "Number of digits in 456 is " << count_digits(456);
    cout << "\nNumber of digits in 7777 is " << count_digits(7777);
    cout << "\nNumber of digits in 4444444 is " << count_digits(4444444);

    cin.get();
    return 0;
}

I don't know how that is working. I mean, it works correctly and counts the digits in the numbers recursively, but I don't know how. I'm writing it down on paper trying to go through it step by step, but I'm not getting any closer to understanding it, and I think I'm even more confused now that I'm working it out on paper. Can someone explain how this works?

[Stewbond]

Let's look at a single case:

What happens when we call count_digits(1234);?

It does not pass the condition in line 7, but does pass the condition in line 12. Therefore it resolves as:
return 1 + count_digits( 123 );

When we look at 123, it does not pass the condition in line 7, but does pass the condition in line 12, so now
return 1 + ( count_digits(123) ); becomes
return 1 + ( 1 + count_digits(12) );

count_digits(12) does not pass the condition in line 7, but does pass the condition in line 12 so now:
return 1 + ( count_digits(123) ); becomes
return 1 + ( 1 + ( count_digits(12) ) ); which becomes
return 1 + ( 1 + ( 1 + count_digits(1) ) );

count_digits(1) passes the condition in line 7, so it returns 1 immediately. so now:
return 1 + ( count_digits(123) ); becomes
return 1 + ( 1 + ( count_digits(12) ) ); which becomes
return 1 + ( 1 + ( 1 + count_digits(1) ) ); which becomes:
return 1 + ( 1 + ( 1 + 1 ) );

The answer is 4.

[Marcos Modenesi]

Nice explanation! You made totally clear.

[cspctec]

Thanks for the explanation, its making more sense now.

[Stewbond]

You're very welcome