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How to determine the root recursive call?

Let's say we write a simple recursive function fib(n)that calculates the nth Fibonacci number. Now we want the function to print this nth number. Since the same function is called repeatedly, there must be a condition that allows only the root call to be printed. The question arises: how to write this condition without passing any additional arguments or using global / static variables.

So we are 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 is different from all the children, returning to another caller, namely the main method in this example. I wanted to know if this property can be used to record a condition and how.

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+5
5

, , - :

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;
}

(++):

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;
}
+5

- , , main. , (- ):

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 main() {
    cout << fib(5) << endl;
    return 0;
}
+5

, , :):

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

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

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