When using array as a parameter, how do get the length of that array parameter in C

Question

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

void insertion_sort(int arr[]);

void insertion_sort(int arr[]) {

    int hold;
    int key;

    for (int i = 2; i < 7; i++) {
        key = arr[i];
        hold = i- 1;
        while (hold >= 0 && arr[hold] > key) {
            arr[hold + 1] = arr[hold];
            hold--;
        }
        arr[hold + 1] = key;
    }

}


int main() {


    int arr[] = {3,4,5,6,7,1,4};

    insertion_sort(arr);

    for (int i = 0; i < sizeof(arr) / sizeof(int); i++) {
        printf("%d", arr[i]);
    }

    return 0;
}

It seems like I cannot use [ sizeof(arr) / sizeof(int) ] in order to get the length of the array in insertion_sort. So I used the integer number instead. What is the reason and what's the proper way of manipulating the array that's taken as a parameter of a function??

score 2 · Answer 1 · answered Mar 08 '17 at 12:45

2

You cannot get the array size inside a function.

What you do instead is to pass the size as another parameter.

void insertion_sort(int arr[], int size)

answered Mar 08 '17 at 12:45

Rishikesh Raje

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score 2 · Answer 2 · answered Mar 08 '17 at 12:49

You can't find the size of an array from inside the function. In the call, the array decays into a pointer to the first element, and clearly a pointer has no concept of how much data it points at in terms of number of elements.

This is simply how C works.

You should add a second argument to indicate the length, and make it have type size_t since that type conveys that we're talking about a size.

See, for instance, the standard library's qsort() function:

void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *));

Basically base and nmemb describe an array.

When using array as a parameter, how do get the length of that array parameter in C

2 Answers2