Is this a good way to free memory?

Question

The function for freeing an instance of struct Foo is given below:

void DestroyFoo(Foo* foo)
{
    if (foo) free(foo);
}

A

Answer 1

What your collegue suggests will make the code "safer" in case the function is called twice (see sleske comment...as "safer" may not mean the same for everybody...;-).

With your code, this will most likely crash:

Foo* foo = malloc( sizeof(Foo) );
DestroyFoo(foo);
DestroyFoo(foo); // will call free on memory already freed

With your collegues's version of the code, this will not crash:

Foo* foo = malloc( sizeof(Foo) );
DestroyFoo(&foo);
DestroyFoo(&foo); // will have no effect

Now, for this specific scenario, doing tmpFoo = 0; (within DestroyFoo) is enough. memset(tmpFoo, 0, sizeof(Foo)); will prevent crash if Foo has extra attributes that could be wrongly accessed after memory is released.

So I would say yes, it may be a good practice to do so....but it's only a kind of security against developers who have bad practices (because there's definitely no reason to call DestroyFoo twice without reallocating it)...in the end, you make DestroyFoo "safer" but slower (it does more stuff to prevent bad usage of it).