CODEWITHSUNDEEP
cpointersmemory-management
curryage
curryage

Reputation: 501

What does free do to a pointer passed by value to a function?

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 ?

Upvotes: 5

Views: 7897

Answers (5)

Germann Arlington
Germann Arlington

Reputation: 3353

What everybody is saying is that your memory will be freed by free(p);, but your original pointer (which you use to call the function with) will still hold the (now invalid) address. If a new block of memory including your address is allocated at a later stage than your original pointer will become valid (for memory manager) again, but will now point to completely different data causing all sorts of problems and confusion.

Upvotes: 1

Sunil Bojanapally
Sunil Bojanapally

Reputation: 12688

Passing p to func() by value, which will copy the pointer and creates the local copy to func() which frees the memory. func() then sets it's own instance of the pointer p to NULL but which is useless. Once the function is complete the parameter p come to end of existence. In calling function you still have pointer p holding an address, but the block is now on the free list and not useful for storage until allocated again.

Upvotes: 2

Pavel Radzivilovsky
Pavel Radzivilovsky

Reputation: 19104

The only problem is if this function is in a different DLL (Windows). Then, it may be linked with a different version of the standard library and have different ideas on how the heap is built.

Otherwise no problem.

Upvotes: 2

benjarobin
benjarobin

Reputation: 4487

No you really free the block of memory. After the function call, the pointer passed to this function is pointing to nowhere : same address but the MMU don't know anymore what to do with this address

Upvotes: 0

user529758
user529758

Reputation:

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

Upvotes: 23

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