Question

I am curious why I am getting the following behaviour in my code.

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


int main(int argc, char *argv[])
{
    int M=24;
    int arr[M];
    int N=24;
    int* ptr=(int*) malloc(sizeof(int)*N); /*Allocate memory of size N  */

    printf("Size of your malloced array is %lu\n",sizeof(ptr)/sizeof(ptr[0])); /* Get the size of memory alloctaed. Should be the same as N?*/

    printf ("Size of your normal arrays is %lu\n",sizeof(arr)/sizeof(arr[0])); /* Ditto  */

    free(ptr);

    return 0;
}

The output is

Size of your malloced array is 2
Size of your normal arrays is 24

I would have thought the output would be 24 in both places. How then does one get the size of the malloced array If somehow I have “forgotten” it?

Surely the pointer ptr will contain some information about the size of the malloced array since when we call free(ptr) it will release the array just malloced

Answer 1

When you use sizeof() on a pointer, you get the size of the pointer. Not the size of the allocated array. In your case, a pointer is probably 8 bytes and an int is 4 bytes, hence why you get 2.

In short, you can’t get the size of an allocated array. You need to keep track of it yourself.

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EDIT : Note that some compilers do actually support this functionality as an extension:

For example, MSVC supports _msize(): http://msdn.microsoft.com/en-us/library/z2s077bc.aspx