How Does NSZombies Really Work?
-
26-06-2021 - |
문제
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?
해결책 2
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.
다른 팁
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
)