The length of array changes after passing it a function

Question

I use the following code to pass an array to a function and print its size before and after calling the function:

#include <stdio.h>

int passing_array(int array[]) {
    int size = sizeof(array);
    printf("The size after passing: %d\n", size);
}

int main () {
    int a[] = {1, 2, 3};

    int size = sizeof(a);
    printf("The size before passing: %d\n", size);

    passing_array(a);

    return 0;
}

After running the code, I got a strange result:

The size before passing: 12

The size after passing: 8

My question is: why does the size of the array change from 12 to 8?

Because the array name decays to a pointer when passing as function argument. — Yu Hao, Sep 16 '15 at 02:32

score 5 · Accepted Answer · answered Sep 16 '15 at 02:32

5

int passing_array(int array[]) {
    int size = sizeof(array);
    printf("The size after passing: %d\n", size);
}

is the same as

int passing_array(int* array) {
    int size = sizeof(array);
    printf("The size after passing: %d\n", size);
}

Hence, size is the size of a pointer.

answered Sep 16 '15 at 02:32

R Sahu

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Just to complement, the first call returns 12 because the size of int in that machine is 4 bytes so and array of 3 elements occupy 12 bytes. – Mandrill Sep 16 '15 at 02:36
Thank you very much for answering my question. – nomorewords Sep 16 '15 at 03:25
@YuanTian, you are welcome. Happy programming. – R Sahu Sep 16 '15 at 03:30

score 2 · Answer 2 · answered Sep 16 '15 at 02:33

2

When you pass an array to a function, it decays to a pointer to the first element of the array. So int array[] in your parameter list is really int *array, and sizeof is returning the size of the pointer.

answered Sep 16 '15 at 02:33

dbush

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Thank you very much, too. But I can only accept one answer, sorry. – nomorewords Sep 16 '15 at 03:25

The length of array changes after passing it a function

2 Answers2