Question about free()

[Mr_Miguel]

When I look at free()'s manpage, it mentions that its behaviour is undefined if I pass it a pointer which hasn't been returned by malloc(), calloc() or realloc(), or if the pointer I pass to it had the memory it's pointing to "free()'d" before.

What if I pass it a pointer which is a copy of a pointer returned by, say, malloc()? For example:

(...)

void free_int_pointer(int* ptr) {
    free(ptr);
}

int main() {
    int* my_pointer = malloc(sizeof(int));
    free_int_pointer(my_pointer);
    return 0;
}

Is this OK to do, even though the pointer that free() receives is a copy of the original "my_pointer"?

Thank you in advance.

[vart]

you should pass the same value that you received from malloc.

As in your sample - you pass the exactly the same value (and do it exactly once) - it is ok

[matsp]

Yes, as vart says, it's the value of the pointer (or put another way, the address you get back from malloc) that is referred to here, not the variable in which it is stored.

So if you get 0x12345600 back from malloc(), you need to free the same address 0x12345600, but where that is stored is up to you, and if you want to shift it around in 15 different variables before you free it, that's also up to you.

You also should not free memory that didn't come from malloc in the first place, e.g.

char *ptr = "Some string";
...
free(ptr);

will most likely crash the system - or cause other undefined behavior.

--
Mats

[pankaj401]

You should set the my_pointer to NULL so that you do not have any dangling references anymore once you free().

[vart]

You should set the my_pointer to NULL so that you do not have any dangling references anymore once you free().

Will not work if you have a copy of this pointer somethere else... For example in the calling function.