Using a function with an array as input in C

Question

I'm trying to write a program which asks the user to insert a series of number to make an array and then, using a function, calculate the sum and the mean of the elements of the array. The problem is that when I use the function in the main the output of sum and mean are both 0. And if I just use arr[n] without & I got other errors, why is that?

Here's the code:

#include <stdio.h>

int sum_and_mean (int arr[], int n);

    int i;
    int n;
    int arr[100];

int main(){

    printf("How many numbers do you want to insert?");
    scanf("%d", &n);

    if((n > 0) && (n <= 20)){
        for(i = 0; i < n; i++){
            printf("Insert a number:");
            scanf("%d", &arr[i]);
        }

        printf("The array is:\n");
        for (i = 0; i < n; i++){
            printf("%d\n", arr[i]);
        }
    }

    else {
        printf("Error: number must be between 0 and 20");
    }

    sum_and_mean(&(arr[n]), n);

    return 0;
}

int sum_and_mean (int arr[], int n){
    double sum = 0;
    double mean;
    for (i = 0; i < n; i++){
        sum += arr[i];
    }
    printf("The sum of the elements is %lf\n", sum);
    mean = sum / n;
    printf("The mean of the elements is %lf", mean);
    return 0;
}

Some programmer dude · Accepted Answer · 2020-04-23 14:16:41Z

4

The expression &(arr[n]) is a pointer to the n:th element in the array, which is outside of the initialized portion of the array.

You want to pass a pointer to the first element of the array: &arr[0], which can also be expressed as plain and simple arr (arrays naturally decays to pointers to their first element).

answered Apr 23, 2020 at 14:16

Some programmer dude

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6 Comments

matteo martelli Over a year ago

I would add that the sum and the mean evaluated by the sum_and_mean function both equal to zero likely because the memory you are accessing from the nth element to the nth + n element is full of zeros. I suggest to try the same code with a smaller array, let's say big about as n itself, to figure out how worse the outcome can get.

Schiele Over a year ago

Why just passing a pointer to the first element of the array is sufficient?

Hitokiri Over a year ago

@Schiele because &arr[0] is exactly arr.

stark Over a year ago

Accessing an array element in C calculates the offset from the first element, so knowing that address is al you need to find the elements.

Some programmer dude Over a year ago

@Schiele And a little tip: If you're struggling with arrays, try drawing it out using pen and paper. Draw the array as a long rectangular box, and divide it into squares, where each square represent an element in the array. Then draw arrows to elements which represents pointers. If you do that for your array, and draw an arrow for the n:th element, then remembering that indexing adds to that will make it quite clear why passing that pointer would not work.

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Hitokiri · Accepted Answer · 2020-04-23 14:30:33Z

1

I see something maybe have to improve in your code (I do not say about using of arr because @Some_programmer_dude explained very clearly).

else {
        printf("Error: number must be between 0 and 20");
    }

    sum_and_mean(&(arr[n]), n);

you should return 0 in else statement because, if you have error, you do not need to call sum_and_mean function.

OT, you define int array, so sum is integer also. You should change double sum to:

int sum;

Thus, for mean value, you can multiply by 1.0 as below if you define int sum:

mean = 1.0 * sum/n

edited Apr 23, 2020 at 14:30

answered Apr 23, 2020 at 14:26

Hitokiri

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2 Comments

Schiele Over a year ago

If I change sum to int then I don't have an exact value of mean

Hitokiri Over a year ago

you can cast int to double by multiplying by 1.0

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