Question

It is known that if we pass a pointer by value to a function, it cannot be freed inside the function, like so:

void func(int *p)
{
    free(p);
    p = NULL;
}

p holds a copy of a (presumably valid) address, so free(p) tries to, well, free it. But since it is a copy, it cannot really free it. How does the call to free() know that it cannot really free it ?

The code above does not produce an error. Does that mean free() just fails silently, “somehow” knowing that address passed in as argument cannot be worked upon ?

Answer 1

p holds a copy of a (presumably valid) address, so free(p) tries to, well, free it. But since it is a copy, it cannot really free it.

It’s not true. free() can work just fine if p is a valid address returned by malloc() (or NULL).

In fact, this is a common pattern for implementing custom “destructor” functions (when writing OO-style code in C).

What you probably mean is that p won’t change to NULL after this – but that’s natural, since you’re passing it by value. If you want to free() and null out the pointer, then pass it by pointer (“byref”):

void func(int **p)
{
    if (p != NULL) {
        free(*p);
        *p = NULL;
    }
}

and use this like

int *p = someConstructor();
func(&p);
// here 'p' will actually be NULL