Why my free() wrapper function doesn't work?

Question

Why should i use pointer to pointer in my saferFree(void**) function when i want to free memory inside that function? Why below code does'n free ip pointer?

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

void saferFree(void *);

int main()
{
    int *ip;
    ip = (int*) malloc(sizeof(int));
    *ip = 5;

    saferFree(ip);

    return 0;
}

void saferFree(void *pp){
    if((void*)pp != NULL){
        free(pp);
        pp = NULL;
    }
}

Above code doesn't free pp pointer but below code with double pointer works properly.I want to know Why?

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



void saferFree(void **);

int main()
{
    int *ip;
    ip = (int*) malloc(sizeof(int));
    *ip = 5;

    saferFree(&ip);
    printf("%p", ip);

    return 0;
}

void saferFree(void **pp){


    if(*pp != NULL){
        free(*pp);
        *pp = NULL;
    }
}

Please show the part where you get to see `.It doesn't Free the allocated memory.` Some people are getting confused. — Sourav Ghosh, Jul 28 '16 at 08:11
I suspect the problem is OP expects the pointer `ip` main() to be `NULL` because it's set to `NULL` in `safeFree()`. `pp` is local to `safeFree()`, so it doesn't reflect in main(). See this post as well: http://stackoverflow.com/questions/13431108/changing-address-contained-by-pointer-using-function — P.P, Jul 28 '16 at 08:15

score 0 · Answer 1 · edited May 23 '17 at 10:33

0

FWIW, it is perfectly safe to pass NULL to free(). Your wrapper is basically meaningless.

Quoting C11, chapter §7.22.3.3, (emphasis mine)

The free function causes the space pointed to by ptr to be deallocated, that is, made available for further allocation. If ptr is a null pointer, no action occurs. [..]

That said, as per your question, it works fine, but because accessing free-d memory is UB, you should not be doing that (which you seem to be doing in main() after calling saferFree() to check if the memory is freed or not. ). Also, maybe worthy to mention that the function arguments are passed by value in C, so any change you make to pp itself (example: pp = NULL;) will not reflect to ip in main(). free() will work normally, though.

Also, please see this discussion on why not to cast the return value of malloc() and family in C..

edited May 23 '17 at 10:33

Community

1
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answered Jul 28 '16 at 08:04

Sourav Ghosh

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3

This seems like a comment for me. – MikeCAT Jul 28 '16 at 08:06
2

Where's freed memory accessed? – P.P Jul 28 '16 at 08:07
@MikeCAT Wait for the OP to edit the question. :) – Sourav Ghosh Jul 28 '16 at 08:08
@usr for a hint, see the comparison in the code block in `main()` as mentioned by OP. – Sourav Ghosh Jul 28 '16 at 08:08
@usr Seen and understood. OP is checking the memory is actually freed or not by attempt to access the memory. – Sourav Ghosh Jul 28 '16 at 08:11
1

@SouravGhosh that isn't said anywhere, OP could be just checking if pointer is NULL – Markus Laire Jul 28 '16 at 08:12
@MarkusLaire The problem is, everyone is very keen on finding secondary issues with the question and answer, while the primary point is, the whole wrapper thing OP is trying to code, is rather useless. Once that is removed, the problem is gone.... :) – Sourav Ghosh Jul 28 '16 at 08:17
`Why my free() wrapper...`. Stop, you don't need one, or least not _this_ one. – Sourav Ghosh Jul 28 '16 at 08:17

Why my free() wrapper function doesn't work?

1 Answers1