Question

Tried the following code :

#include<stdio.h>
int main()
{        
        int *p,*q;
        p = (int *)malloc(sizeof(int));
        *p =10;
        q = p;
        printf("%u \n",p);
        printf("%u \n",q);
        free(p);
        printf("%u \n",p);
        return 0;
}

The output got is as follows :

[root@lnxdesk Tazim]# ./a.out
154804232
154804232
154804232

Why is that address inside p is still printed even if I have done free(p);?
What has free(p) done then?

I want to understand the concept of free/malloc clearly. Any help will be valuable.

Answer 1

free() only frees the memory on the heap. It does not change the value of your pointer. If you tried to print the memory pointed by your pointer, you’ll probably get some kind of garbage.

Also, when you called free, you gave it the pointer, not the address to your pointer, so free can’t change your pointer…