How to detect the root recursive call?

Question

Say we're writing a simple recursive function fib(n) that calculates the nth Fibonacci number. Now, we want the function to print that nth number. As the same function is being called repeatedly, there has to be a condition that allows only the root call to print. The question is: how to write this condition without passing any additional arguments, or using global/static variables.

So, we're dealing with something like this:

int fib(int n) {
    if(n <= 0) return 0;
    int fn = 1;
    if(n > 2) fn = fib(n-2) + fib(n-1);
    if(???) cout << fn << endl;
    return fn;
}

int main() {
    fib(5);
    return 0;
}

I thought that the root call differs from all children by returning to a different caller, namely the main method in this example. I wanted to know whether it is possible to use this property to write the condition and how.

Update: please note that this is a contrived example that only serves to present the idea. This should be clear from the tags. I'm not looking for standard solutions. Thanks.

Answer 1

The way I would do this is with a helper function:

int _fib(int n) {
    if(n <= 0) return 0;
    int fn = 1;
    if(n > 2) fn = _fib(n-2) + _fib(n-1);
    return fn;
}

int fib(int n) {
    int fn=_fib(n);
    cout << fn << endl;
    return fn;
}

Alternatively, you could write it like this (c++):

int fib(int n, bool f=true) {
    if(n <= 0) return 0;
    int fn = 1;
    if(n > 2) fn = fib(n-2, false) + fib(n-1, false);
    if(f) cout << fn << endl;
    return fn;
}