Passing arrays to functions

Learn to pass arrays to functions as pointers.

Passing arrays to functions

When you pass an array to a function the array name decays into a pointer. In the function signature you can write int arr[] or int *arr; both mean the same.

int sum(int arr[], int n) {
    int total = 0;
    for (int i = 0; i < n; i++) total += arr[i];
    return total;
}

The array length does not come with the pointer; the n parameter is passed separately. The function sees the original array, not a copy.

sizeof gives pointer size on a parameter

Inside a function sizeof(arr) does not give the array size; it gives the pointer size (usually 4 or 8 bytes). This is why you must pass the array length as a separate parameter.

void print_all(int *arr, int n) {
    /* sizeof(arr) is 4 or 8 here! */
    for (int i = 0; i < n; i++) printf("%d ", arr[i]);
}

This is the most common pitfall of array decay. Always carry the length separately.

Exercise Write a sum function

Define int sum(int *arr, int n): with a for loop sum the array elements and return the total variable. In main define int arr[3] = {4, 5, 6}; and print the result with printf("%d\n", sum(arr, 3));. Output should be 15.

Show example solution

#include <stdio.h>

int sum(int *arr, int n) {
    int total = 0;
    for (int i = 0; i < n; i++) {
        total += arr[i];
    }
    return total;
}

int main(void) {
    int arr[3] = {4, 5, 6};
    printf("%d\n", sum(arr, 3));
    return 0;
}

Test yourself

Match signatures and meanings

int arr[]

array parameter
int *arr

pointer parameter
arr, 3

array and size
sum(arr, n)

function call
Which signature accepts an int array?
- void f(int[] arr)
- void f(int arr)
- void f(int *arr[])
- void f(int arr[])
Show answer

void f(int arr[]) — int arr[] accepts an int array (as a pointer).

Pass array and size to function

int arr[3] = {1, 2, 3};
sum(arr, ____);  /* pass 3-element array */

Show answer

int arr[3] = {1, 2, 3};
sum(arr, 3);  /* pass 3-element array */

Next lesson: Pointer loop

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