Fibonacci One Recursive Call

Question

I wish to transform the basic Fibonacci function:

int find_fib(int fib_to) {

    if (fib_to == 0) {
        return 0;
        
    }
    else if (fib_to == 1) {
        return 1;
        
    }
    else {
        return (find_fib(fib_to - 1) + find_fib(fib_to - 2));       
    }   
}

into one that would use only ONE recursive call. I searched up many sites that tells me to store the value found by (find_fib(fib_to - 1) + find_fib(fib_to - 2)) into some array and then make use of the array, but doing so requires 2 recursive calls.

Any tips on how to solve the problem?

Answer 1

You mean something like that?

#include <stdio.h>

int fibonacci(int number_iterations, int number_a, int number_b) 
{
    if(number_iterations <= 0) {
        printf("error: 'number_iterations' must be >= 1\n");
        return -1;
    }

    if(number_iterations == 1) {
        printf("%d ", number_a + number_b);
        return number_a + number_b;
    }

    printf("%d ", number_a + number_b);
    return fibonacci(number_iterations - 1, number_b, number_a + number_b);
}

int main()
{
    int a = 1; 
    int b = 1;
    int number_of_iterations = 0;

    printf("Enter a number >= 1: ");

    int err = scanf("%d", &number_of_iterations);
    int final_result = -1;

    if (err != EOF) {
        printf("fibonacci: %d %d ", a, b);
        final_result = fibonacci(number_of_iterations, a, b);
    }

    printf("Final fibonacci: %d\n", final_result);

    return 0;
}

would return you:

Enter a number >= 1: 10
fibonacci: 1 1 2 3 5 8 13 21 34 55 89 144
Final fibonacci: 144