C - Array sum function constantly returning zero

Question

I'm trying to write a a program that lets the user generate anywhere from 1-75 Fibonacci numbers, print them, and then add all of those values together to get to over all sum. I had no problem generating the numbers and printing them, but for some reason I'm having a lot more trouble with the sum function. For some reason it only returns 0...and after staring at this for as many hours as I have I'm at a loss. Any and all help would be greatly appreciated. My code is below:

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

//function to calculate fibonacci number 1-75 (user specified)
int fibonacci(int * array, int size)
{
    int i;
    unsigned long long int Fibonacci[size]; //generating the array to save the Fibonacci values

    Fibonacci[0] = 0; //by definition
    Fibonacci[1] = 1; //by definition

    for (i = 2; i < size; i++) {
        Fibonacci[i] = Fibonacci[i-2] + Fibonacci[i-1]; //equation to calculate the numbers
    }

    for (i = 0; i < size; i++) {
        printf("%llu\t", Fibonacci[i]); //printing out the numbers
    }

    printf("\n");
    return size;
}

int fibonacciSum(int * array, int size)
{
    int i;
    int sum = 0;

    for (i = 0; i < size; i++) {
        sum += array[i];
    }

    return sum;
}

int main(void)
{
    int size;

    printf("How many Fibonacci numbers do you want to generate (between 1 and 75)?\n");
    scanf("%i", &size);

    int * FibonacciNumbers = (int*) calloc(size,sizeof(int));

    if (size < 1 || size > 75) {
        printf("Bad number!");
        return 1;
    }

    fibonacci(FibonacciNumbers, size);
    printf("The sum of the Fibonacci numbers is: %d\n", fibonacciSum(FibonacciNumbers, size));

    return 0;

}

Deck · Accepted Answer · 2013-11-19 03:08:51Z

6

If you rewrite function fibonacci as

int fibonacci(int * array, int size)
{
    int i;

    array[0] = 0; 
    array[1] = 1; 

    for (i = 2; i < size; i++) {
        array[i] = array[i-2] + array[i-1]; //equation to calculate the numbers
    }

    for (i = 0; i < size; i++) {
        printf("%d\t", array[i]); //printing out the numbers
    }

    printf("\n");
    return size;
}

Because your job on Fibonacci array doesn't affect on array which is stored in main and used in other functions.

answered Nov 19, 2013 at 3:08

Deck

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randomusername · Accepted Answer · 2013-11-19 03:01:40Z

2

The Fibbonacci array is only accessible from within the function. You need to store the numbers in another array.

answered Nov 19, 2013 at 3:01

randomusername

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3 Comments

None Over a year ago

That "other array" will be the passed parameter which is currently unused.

Eric Over a year ago

correction: @Roger is correct... you're passing in a perfectly well-allocated array to contain the results, but you're ignoring it and instead using an array locally allocated within the fibonacci function which will be tossed as soon as the function returns.

ImpurestClub Over a year ago

Wow...thank you. I knew it was something dumb that I just couldn't see. This fixed it.

xd6_ · Accepted Answer · 2013-11-19 03:08:44Z

0

forgetting to assign ? just not sure why the first one returns anything.

size = fibonacci(FibonacciNumbers, size);

And you mixed up the parameter name inside the function.

answered Nov 19, 2013 at 3:08

xd6_

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