How Does NSZombies Really Work?

Question

I cannot find any detailed apple documentation on how the NSZombie really functions. I understand that its designed to not actually release objects and just maintain a count of references to catch any extra releases, but how would something like this work:

for(int i = 1; i < 10; i++)
{
NSMutableArray *array = [[NSMutableArray alloc] initWithCapacity: i];

[array release];

}

Since the same variable/object is being allocated/initialized and released in the same application, how would NSZombie's technically handle this? I know that this shouldn't flag any zombies because every alloc has a release, but how would Xcode technically handle re-allocating the same memory with different capacities?

Answer 1

This question was answered in the comments by Brad Larson.

Quote:

That isn't the same object, or the same memory. You're creating a distinct, new NSMutableArray instance on every pass through that loop. Just because a pointer to each is assigned to array does not make them the same object.

A pointer merely points to a particular location in memory where the object exists. A given object in memory can have multiple pointers to it, or even none (when it is being leaked). NSZombie acts on the object itself, not pointers to it.

Answer 2

With Zombies, objects don't actually need to be freed[1] -- the object is simply turned into a "Zombie" at some point after the object's retain count reaches 0. When you message a "Zombified" instance, a special error handler is performed.

1) Freeing Zombies is optional. Unless you really need the memory for a long running or memory intensive task, it is a more effective test to not to free the zombies (NSDeallocateZombies = NO)