Question

I have this code:

#include <stdio.h>
#include <stdlib.h>


int func(int n0, int n);

int main ()
{
    int n0, n, nFinal=0;
    printf ("Enter constant (n0): ");
    scanf ("%d", &n0);
    printf ("Enter the number of iteractions (n): ");
    scanf ("%d", &n);
    nFinal = func(n0, n);
    printf ("nFinal after %d iteractions is %d: \n", n, nFinal);
    return 0;
}

int func(int n0, int n){
    int i,nFinal=0;

    for (i = 0; i < n; i++){
        nFinal = (nFinal*nFinal) + n0;
    }

    return nFinal;
}

The nFinal is calculated within a for loop. I would like to achieve the same result but doing a recursive function.

From what I see, I cannot change the function call because I always need the start number and the number of iterations. Therefore, after the first iteration, the program would have to call again nFinal = func (n0, n); but as I need at each iteration of the calculated value of nFinal, I will have to change this.

Is it possible to make a recursive function but maintaining the function as nFinal = func (n0, n);?

Can someone point me some way?

Answer 1

Looking at your function

int func(int n0, int n){
    int i,nFinal=0;

    for (i = 0; i < n; i++){
        nFinal = (nFinal*nFinal) + n0;
    }

    return nFinal;
}

If n is (less than) 0, your result is 0. Then the new value of nFinal is the old value of nFinal^2 + n0, so you get:

int func(int n0, int n){
    if (n <= 0) return 0;

    int f = func(n0, n-1);
    return f*f + n0;
}