Why the size of malloc-ed array and non-malloced array are different?

Question

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

int main (int argc, char *argv[]) {

    int* arr1 = (int*)malloc(sizeof(int)*4);
    int arr2[4];


    printf("%d \n", sizeof(arr1));
    printf("%d \n", sizeof(arr2));

    free(arr1);

    return 0;
}

Output

8
16

Why?

Perhaps a better duplicate: http://stackoverflow.com/questions/492384/how-to-find-the-sizeofa-pointer-pointing-to-an-array — Lundin, Sep 12 '16 at 07:02

Sourav Ghosh · Accepted Answer · 2016-09-12T06:21:50.800

3

Arrays are not pointers.

In your code, arr1 is a pointer, arr2 is an array.

Type of arr1 is int *, whereas, arr2 is of type int [4]. So sizeof produces different results. Your code is equivalent to

sizeof (int *);
sizeof (int [4]);

That said, sizeof yields the result of type size_t, so you should be using %zu to print the result.

edited Sep 12 '16 at 06:21

answered Sep 12 '16 at 06:13

Sourav Ghosh

133,132
16
183
261

How to measure size of heap, what was malloc pointer on heap? – Jeungwoo Park Sep 12 '16 at 06:31
@JeungwooPark sadly, you cannot directly know the size of the allocated memory from the returned pointer, you have to keep track of yourself. – Sourav Ghosh Sep 12 '16 at 06:38

score 0 · Answer 2 · answered Sep 12 '16 at 06:15

0

arr1 is a pointer when you call sizeof(arr1) this will you size of pointer. but arr2 represent block of memory for 4 integers.

answered Sep 12 '16 at 06:15

Abhishek

379
1
8

Why the size of malloc-ed array and non-malloced array are different?

2 Answers2

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