Objective-C pointers/memory management question - iphone

I am testing the following code below. ffv is declared in the interface file.
ffv = [[FullFunctionView alloc] initWithFrame:self.view.bounds];
NSLog(#"%i", [ffv retainCount]); // prints 1
[self.view insertSubview:ffv belowSubview:switchViewsBtn];
NSLog(#"%i", [ffv retainCount]); // prints 2
[ffv release]; // you can release it now since the view has ownership of ffv
NSLog(#"%i", [ffv retainCount]); // prints 1
if (ffv == nil)
NSLog(#"ffv is nil");
// "ffv is nil" is not printed
[ffv testMethod]; // "test method called" is printed
this is my [ffv testMethod] implementation
- (void)testMethod
{
NSLog(#"test method called");
}
What I deduce in this case is that even if you release an object with retain count 2, you lose ownership of that object however, the reference is still kept.
Now, my question are:
Is my deduction correct?
Is there anything else important that can be deduced from this?
What are the complications caused by still keeping (using) ffv and calling methods from ffv? (My opinion is that this is ok since the view will always own ffv and won't release it until someone calls viewDidUnload. And as long as I don't pass ffv's reference to other objects.)

There are a couple of problems with using ffv after you have released it and it's only retained by your view controller's view.
1) It introduces a potential for future bugs, because later you might not remember that ffv is otherwise not retained. When you release the view (e.g. by replacing it with another view), you have a dangling pointer that you still hold a reference to.
2) In the special case of a UIViewController the view could be released at any time (you usually never call viewDidUnload yourself). The default behavior of UIViewController, when receiving a memory warning and the view is currently not visible, is to release the view, so unless you set the reference to nil in viewDidUnload, you have a dangling pointer again, even though you never explicitly released the view yourself.

1) Is my deduction correct?
Your deduction is correct. The Memory Management Programming Guide explains that each object has one or many owners. You own any object you create using any method starting with alloc, new, copy, or mutableCopy. You can also take ownership of an object using retain. When you're done with an object, you must relinquish ownership using release or autorelease.
Releasing the object doesn't change the value of any variables that reference that object. Your variable contains the object's memory address until you reassign it, no matter what retain count the object has. Even if the object's retain count goes to zero, causing the object to get deallocated, your variable will still point at that same address. If you try to access the object after it's been deallocated, your app will normally crash with EXC_BAD_ACCESS. This is a common memory management bug.
2) Is there anything else important that can be deduced from this?
Nothing comes to mind.
3) What are the complications caused by still keeping (using) ffv and calling methods from ffv? (My opinion is that this is ok since the view will always own ffv and won't release it until someone calls viewDidUnload. And as long as I don't pass ffv's reference to other objects.)
When you call release, you are telling the Objective C runtime that you no longer require access to the object. While there may be many cases like this one in which you know the object will still exist, in practice you really shouldn't access an object after calling release. You'd just be tempting fate and setting yourself up for future bugs.
I personally don't like peppering my code with release statements, because I don't trust myself to remember them 100% of the time. Instead, I prefer to autorelease my variables as soon as I allocate them like this:
ffv = [[[FullFunctionView alloc] initWithFrame:self.view.bounds] autorelease];
This guarantees that ffv will exist at least until the end of the method. It will get released shortly thereafter, typically before the next iteration of the run loop. (In theory this could consume excessive memory if you're allocating a large number of temporary objects in a tight loop, but in practice I've never encountered this case. If I ever do, it will be easy to optimize.)

The object is not deallocated until the retain count goes to 0. As long as it's not deallocated, you can keep using it without trouble. By retaining it you ensure that it won't be deallocated under your feet; however, if you retain another object that you know retains the first object, you can get away with this form of "indirect retaining". Don't complain when you move things around later and things start breaking, though.

if (ffv == nil)
NSLog(#"ffv is nil");
// "ffv is nil" is not printed
That's correct, releasing an object does not set the pointer to nil even if it is dealloced at that time. Good practice is to always set your pointer to nil after you release it.
You are correct to say that after you released it, it wasn't dealloced because the view still had a retain on it. But that's not how you should be thinking about it. If you want to use that object and you want it to be alive, retain it. Doesn't matter who else is retaining it. Your object has nothing to do with those other objects. You want it, retain it. You're done with it, release it and set your pointers to nil. If you don't set it to nil and everyone else also released it, you will have a dangling pointer to an object that was dealloced, and that will cause you a crash and much grievance.
so this:
[ffv release]; // you can release it now since the view has ownership of ffv
ffv = nil; // you released it, so that means you don't want it anymore, so set the pointer to nil
if you still want to use it, don't release it until you're done with it.

Well, I'm not sure 'losing' ownership is the right term. In Objective-C you have to carefully marshal your ownership of the object. If you create or retain an object, you are responsible for releasing it (either directly or via an autorelease pool). When you call release however, you don't lose a reference to the object, if something else has retained it, it will still be in memory, and your pointer will still potentially point to it.
You have a pointer ffv which is just a pointer to some memory, and you have the object which is created in the first line that ffv points to.
By calling release, you are stating that you no longer require the ponter ffv to point to a valid object, that in this context you would be happy for the object to be deallocated. The pointer still points to that bit of memory, and it is still there because its retain count was increased by assigning it to the view.
The line [ffv testMethod] is in danger of not working, as it follows the release and may not point to a valid object. It only works because something else is keeping it alive. ffv still has the same address value that it had when it was first assigned.
So in order:
Your deduction is correct.
Not really.
You shouldn't use ffv after the release call. You have no guarantee that the object is going to be there for you.
These are pointers we are using here, not references like you find in Java or C#. You have to marshal your ownership of the object, you create it, have some pointers to it and by careful management of retain and release calls you keep it in memory for as long as you need it.

Related

What happens if you try to release an object that has been released?

What if I have an object, say NSIndexPath in which before I do a copy I always release it first?
Is it possible to have a memory count under 0?
I am doing this to prevent memory leaks.. is this a good way?
//global already has some value before in here or it doesn't have a value.. I want to update
//this with a new value (I no longer care to the old pointer)
[global release]
global = [indexPath copy];
Don't. When the retain count reaches 0, your object will be dealloc'ed and its pointer will become invalid, so using it again will cause unpredictable results (namely crashing).
You should read through Apple's Memory Management Guide.
This is the fundamental rule:
You only release or autorelease objects you own. You take ownership of an object if you create it using a method whose name begins with “alloc”, “new”, “copy”, or “mutableCopy” (for example, alloc, newObject, or mutableCopy), or if you send it a retain message.
You use release or autorelease to relinquish ownership of an object. autorelease just means “send a release message in the future” (specifically: when the used autorelease pool receives a drain message).
Update:
As Josh pointed out, the one case you need to consider is when global and indexPath are the same. In that case, you would still "need" the pointer (to perform the copy), so you either autorelease (instead of releasing) or you use a temporary variable to handle that.
What you are doing is essentially correct, so long as global either has an old value that you no longer need or is nil. If it is one of a class's ivars, it will be nil when an instance of the class is created. The one catch is that if the new indexPath happens to be the same object as the one that's already in global, you will have over-released and you will crash.
// global points to object at 0x8BADFOOD
// indexPath also happens to be 0x8BADFOOD; for some reason, it
// hasn't changed since last time. This _can_ happen.
[global release]; // object at 0x8BADFOOD is deallocated
global = [indexPath copy]; // indexPath no longer valid!
// Crash! Bang! Boom!
The way to avoid this is to use a temp variable.
When you declare a property as copy and synthesize that property, the setter method that is created looks essentially like this, and you can do the same thing:
- (void)setMyFloozit:(Floozit *)newFloozit {
// Copy first in case newFloozit and myFloozit are for
// some reason the same object
// Take ownership of a copy of newFloozit
Floozit * tmp = [newFloozit copy];
// We no longer need old value of myFloozit, so we release it.
[myFloozit release];
// tmp contains a value that we own, we just need to assign
// it to the correct name.
myFloozit = tmp;
}
This could be made slightly better by checking first to see whether newFloozit and myFloozit are the same, and doing nothing if they are.

What happens after I release a deallocated object?

I created
object *Obj = [[Obj alloc] init];
Obj retain count is 1. After I release it, the object is deallocated.
If I try to release the object again, what will happen?
EXT_BAD_ACCESS most likely since your object reference is no longer valid.
The code may crash. But it may just as well work most of the time.
You brake a rule, you may get caught. But you may just as well get away with it, living in constant fear that you may get caught later on.
There’s an important distinction to be made here: you can’t release the object again, because the object no longer exists. If you send another message to the variable (be it release or any other message), the behaviour is undefined because the variable is no longer known to point to a valid object. (It’s possible that the address the variable now points to will have been reused for a different object, in which case it may not crash, but of course that’s still a bug.)
Once the retain count of an object reaches 0, it is released, and all further attempts to access it will result in random behaviour.
If you use autorelease instead, the retain count will not be lowered, and the object will be put in the autoreleasepool. The object will only lower its retain count once it reaches the autoreleasepool drain command, which is usually done on a much higher level in a much broader scope. If you really need the object after the autoreleasepool is drained, you should retain it before drain is executed, or else it will have exactly the same behaviour as in my first paragraph.
Get EXT_BAD_ACCESS. Because of you are already release it and now try to release again.
your object reference is no longer valid.

Any way to check if an instance is still in memory?

Example: I have a view controller and get rid of it. But there's still an variable holding it's memory address. Accessing that results in EXEC_BAD_ACCESS. Of course. But: Is there any way to check if that variable is still valid? i.e. if it's still pointing to something that exists in memory?
You need to read this again:
Cocoa Memory Management Guidelines
In short, if you want something to stick around you must retain it.
If you want something to go away and you have previously retained it, you must release or autorelease it.
You must never call dealloc directly (except [super dealloc]; at the end of every one of your dealloc methods).
You must never release or autorelease an object that you did not retain.
Note that some methods do return retained objects that you must release. If you alloc an instance of a class, that implies a retain. If you copy and instance, the copy is retained.
If you are ever tempted to use the retainCount method, don't. It isn't useful. Only consider retain counts as a delta; if you add, you must subtract, but the absolute value is an implementation detail that should be ignored.
(In other words, even if there were ways to check for an object's validity definitively -- there aren't -- it would be the wrong answer.)
Oh, and use the Build and Analyze feature in Xcode. It does a very good -- but not quite perfect -- job of identifying memory management problems, amongst other things.
That's what the entire memory management model is set up for - if you call retain at the right times, and release and autorelease at the right times, that can't happen. You can use NSZombie to help you debug.
Use "NSZombieEnabled" break point.
For this reason only all strongly recommend us to use accessors. If your object is released anywhere, it will get assigned to nil, and there will be no harm if you call any API or method on Nil object. So please make a habit of using Accessors.
you just add this NSZombieEnabled Flag as an argument to your application in build settings. and enable it. Now you run your application in debug mode. If any such crash is about to occur, this breakpoint will show you which object is freed and where it is crashing.
Cheers,
Manjunath
If by variable, you mean whether the pointer to your object still references valid memory then:
MyClass *myVariable = [[MyClass alloc] init];
//Tons of stuff happens...
if (myVariable != nil)
//Do more stuff

iPhone - dealloc - Release vs. nil

Wondering if someone with experience could possibly explain this a bit more. I have seen examples of...
[view release];
view = nil;
....inside the (void) dealloc.
What is the difference and is one better then the other?
What is the best way?
When doing retainCount testing I have personally seen nil drop a count from 3 to 0 for me, but release only drops it from 3 to 2.
What you have seen is probably these:
1) [foo release];
2) self.bar = nil;
3) baz = nil;
Is releasing the object, accessing it through the instance variable foo. The instance variable will become a dangling pointer. This is the preferred method in dealloc.
Is assigning nil to a property bar on self, that will in practice release whatever the property is currently retaining. Do this if you have a custom setter for the property, that is supposed to cleanup more than just the instance variable backing the property.
Will overwrite the pointer baz referencing the object with nil, but not release the object. The result is a memory leak. Never do this.
If you are not using properties (where self.property = nil will also release an object) then you should ALWAYS follow a release by code that sets the reference to nil, as you outlined:
[view release]; view = nil;
The reason is that it avoids he possibility that a reference can be used that is invalid. It's rare and hard to have happen, but it can occur.
This is even more important in viewDidUnload, if you are freeing IBOutlets - that's a more realistic scenario where a reference might go bad because of memory warnings unloading a view, and then some other code in the view trying to make use of a reference before the view is reloaded.
Basically it's just good practice and it will save you a crash at some point if you make it a habit to do this.
#bbullis22 you have seen the restain count drop from 3 to 0 because you set the reference to nil. then you asked for the retaincount of 'nil' which is zero. however, the object that used to be referenced has the same retain count - 1 (due to setting the reference to nil).
using release, the reference still references the same object, so that's why you see the retain count drop from 3 to 2 in this situation.
As far as usage inside your code, in your dealloc you don't need the assignment to the property, releasing is all you need to do.
- (void)dealloc {
[myProperty release]; // don't need to assign since you won't have the object soon anyway
[super dealloc];
}
I think using both is kind of safety net. With only release in place you could run in problem if you screwed reference counting management. You would release an object, giving its memory back to system but pointer would be still valid.
With nil you are guaranteed that program will not crash since sending message to nil does nothing.

iphone memory management issue

How to check if an object has been released from the memory?
I know a object has to be released manually when we use alloc|copy|retain to create that object. if use the instance class method (NSString stringwithformat:), the object will be released automatically by NSAutoRealeasePool, however, sometime there have some object used to release at the (void)dealloc function there.
Can anybody tell me
What is the difference between [object release] instantly after alloc function, [object release] at (void)dealloc,and NSAutoReleasePool to release the object?
When is the (void)dealloc function being called and when is the NSAutoReleasePool release the memory, I know it is every loop cycle, how can I check it from code?
When I declared a variable in the header and set the property as retain, I still can use the alloc function to that object but not cause a memory leak, but from this post it seems like once declared a retain object at header, then you no longer can alloc the object because both will retain the object,can anybody tell me why I still can code like this without causing any memory leak?
[object release] releases your object immediately. After sending a release message you should not use the object again - unless you absolutely know you still have a retain on that object. If yours was the last retain the memory could be freed during the call to release.
Autorelease frees an object 'sometime later' the system does not guarantee anything about the meaning of 'later' other than that it will be after the scope of the current message.
See above, there is no real qay to guarantee when dealloc is called following autorelease from the point of view of your code. You should just assume it is sometime after the return of the method where you send the autorelease message.
You simply need to balance retain and release. If you have one to many (as is likely i nthe situation you describe) that is a leak. Conversely if you have unbalance the other way, you will generate a more destructive error when you access freed memory.
The correct sequence for a retained property is:
alloc init // retain == 1
set the property // retain == 2 due to setProperty calling retain.
release the object // retain == 1
Which leaves a retain count of one, no memory leak.