Memory management in local variable objective-c - objective-c

In one interview i was asked to implement NSArray's exchangeObjectAtIndex:withObjectAtIndex: method.
I wrote the following code:
- (void)exchangeObjectAtIndex:(NSUInteger)index1 withObjectAtIndex:(NSUInteger)index2 {
id tmp = [self objectAtIndex:index1];
[self replaceObjectAtIndex:index1 withObject:[self objectAtIndex:index2]];
[self replaceObjectAtIndex:index2 withObject:tmp];
}
Interviewer said here's a memory management problem in first line and I'm going to catch bad_access_exc.
He recommended to write as this:
- (void)exchangeObjectAtIndex:(NSUInteger)index1 withObjectAtIndex:(NSUInteger)index2 {
id tmp = [[[self objectAtIndex:index1] retain] autorelease];
[self replaceObjectAtIndex:index1 withObject:[self objectAtIndex:index2]];
[self replaceObjectAtIndex:index2 withObject:tmp];
}
I understand that his code is right, but since tmp is local variable and it's going to be assigned, so there's no releasing and everything is gonna be ok. Is there any error?

If you are using manual memory management, there is an error. Apple has documented the problem under “Avoid Causing Deallocation of Objects You’re Using” in the Advanced Memory Management Programming Guide.
Specifically, objectAtIndex: doesn't retain and autorelease the object that it returns to you. So the NSArray might have the only “owning” reference to the object. Assigning to tmp under manual retain counting (MRC) doesn't retain the object so tmp doesn't own it and the autorelease pool doesn't own it.
This means that when line 2 of your method sends [self replaceObjectAtIndex:index1 withObject:[self objectAtIndex:index2]], the array might release the last reference to the object, deallocating it. At that point, tmp refers to a deallocated object; this is called a “dangling reference”.
Then in line 3, you try to put the dangling reference in the array. The array will send retain to the reference, which is invalid, and you will crash or experience heap corruption.
Under ARC, assigning to tmp does retain the object, so there is no error in that case.

Remember that id tmp is nothing more than a pointer to the object in your array. It doesn't say anything about the memory management of the object it's pointing to.
...it's going to be assigned, so there's no releasing...
This is the sticking point here. You can't guarantee that the object at index1 won't be deallocated when you replace it with the object at index2. In fact, the array will call release on it at this point to balance out the retain it called on the object when it was originally added to the array. Thus, it's possible that when the object at index1 is replaced will the object at index2, the reference count of the object at index1 will go to zero, the object will be deallocated, and your tmp variable will turn into a dangling pointer. The ... retain] autorelease] dance keeps the object around long enough to do the swap without having to worry about it deallocating before the end of the method (likely it will stick around until the top of the next run loop).

Related

ARC and __unsafe_unretained

I think I have a pretty good understanding of ARC and the proper use cases for selecting an appropriate lifetime qualifiers (__strong, __weak, __unsafe_unretained, and __autoreleasing). However, in my testing, I've found one example that doesn't make sense to me.
As I understand it, both __weak and __unsafe_unretained do not add a retain count. Therefore, if there are no other __strong pointers to the object, it is instantly deallocated (with immutable strings being an exception to this rule). The only difference in this process is that __weak pointers are set to nil, and __unsafe_unretained pointers are left alone.
If I create a __weak pointer to a simple, custom object (composed of one NSString property), I see the expected (null) value when trying to access a property:
Test * __weak myTest = [[Test alloc] init];
myTest.myVal = #"Hi!";
NSLog(#"Value: %#", myTest.myVal); // Prints Value: (null)
Similarly, I would expect the __unsafe_unretained lifetime qualifier to cause a crash, due to the resulting dangling pointer. However, it doesn't. In this next test, I see the actual value:
Test * __unsafe_unretained myTest = [[Test alloc] init];
myTest.myVal = #"Hi!";
NSLog(#"Value: %#", myTest.myVal); // Prints Value: Hi!
Why doesn't the __unsafe_unretained object become deallocated?
[EDIT]: The object is being deallocated... if I try to substitute lines 2 - 3 with NSLog(#"%#", myTest); the app crashes (and an overridden dealloc in Test is being called immediately after the first line). I know that immutable strings will continue to be available even with __unsafe_unretained, and that a direct pointer to the NSString would work. I am just surprised that I could set a property on a deallocated object (line 2), and that it could later be dereferenced from a pointer to the deallocated object it belonged to (line 3)! If anyone could explain that, it would definitely answer my question.
I am just surprised that I could set a property on a deallocated object (line 2), and that it could later be dereferenced from a pointer to the deallocated object it belonged to (line 3)! If anyone could explain that, it would definitely answer my question.
When the object is deallocated it is not zeroed. As you have a pointer to the deallocated object and the property value is stored at some offset to that pointer it is possible that storing and retrieving that property value will succeed after deallocation, it is also quite possible that everything will blow up for some reason or other.
That your code works is quite fragile, try debugging it with "Show Disassembly While Debugging" and stepping through, you'll probably hit an access violation, or take down Xcode itself...
You should never be surprised that strange things happen in C, Objective-C, C++ or any of the family; instead reserve your surprise for so few strange things happening!
Because the constant string in objc is a constant pointer to a heap address and the address is still valid.
edited after comment:
Maybe because the memory at the test objects address hasn't been overwritten and still contains that object? Speculating....
You can see when Test is deallocated by implementing its -dealloc method and adding some simple logging.
However, even if Test is deallocated immediately, the memory it occupied in RAM may remain unchanged at the time you call myVal.
#"hi!" produces a static global constant string instance that is, effectively, a singleton. Thus, it'll never be deallocated because it wasn't really allocated in the first place (at least, it really isn't a normal heap allocation).
Anytime you want to explore object lifespan issues, always use a subclass of NSObject both to guarantee behavior and to make it easy to drop in logging hooks by overriding behavior.
Nothing strange there…
You need to have at least 1 strong reference to object to keep it alive.
Test * anTest = [[Test alloc] init];
Test * __weak myTest = anTest;
myTest.myVal = #"Hi!";
NSLog(#"Value: %#", myTest.myVal); // Prints Value: (Hi)

retain release common practice in iOS setter methods

The common practice in IOS setters is the following:
- (void)setMyString:(NSString *)newString {
if ( newString != myString ) {
[myString release];
myString = [newString retain];
}
}
On the contrary this is not good practice
- (void)setMyString:(NSString *)newString {
if ( myString != nil ) [myString release];
myString = [newString retain];
}
}
What is the reason checking for equality in the first case? What is the problem in the seconds case?
If you set something like this [object setMyString:[object myString]]; without checking for equality - it will be crash! Because it will be released before you send it message retain. (in case when only object own string). Also in first example we checking for equality to avoid extra operations.
I know this is somewhat redundant, but...
If the new and old object are the same then you send release to the old object and it gets deallocated, the pointer to the new object will become a dangling pointer as the object it pointed to will no longer exist (since it pointed to the same object as the old object pointer). Ex. if myString and newString point to the same instance who has a retain count of one, then you subtract one, it'll equal zero. it's too late to add one now, because it'll get deallocated. However, reverse the calls to retain and release and it should be fine. If the retain count is one and you add one, it's now two, and you can safely send release. In general, I'd say before you disown an object, assert ownership of the new one first.
Also, the first type of setter would be what you would use for retain/strong style setter. If it were assign you wouldn't need to retain/release as no ownership is supposed to be asserted. NSStrings often have a copy style setter which copies the argument and uses that, which would create a copy instead of retaining. I would generally use copy for anything with a mutable subclass as you wouldn't want someone passing in a NSMutableString and mutating it behind your back. This page goes into accessors, and you'll notice that they retain the new value before releasing the old one, and explain why.
You may take this to though the memory management, https://developer.apple.com/documentation/Cocoa/Conceptual/MemoryMgmt/.

use of [release] in objective C

I have a doubt regarding memory management in Objective-C.
-(void)viewDidLoad
{
NSNumber *num=[[NSNumber alloc] initWithInt:10];
[num release];
NSLog(#”%i”,num);
}
The above code is working fine by printing the value. But as soon as the object has been released it loses its value right? Then how come its working fine?
Here's what's going on. First you do this:
NSNumber *num=[[NSNumber alloc] initWithInt:10];
Your num variable now contains a pointer to an NSNumber object, and (because you used alloc) you own that object. Then you do this:
[num release];
When you sent release to the object, you relinquished your ownership of it. The object might still exist and be unchanged, or it might have been destroyed. You don't know. Your num variable still contains the same pointer, but the memory it points to might not be a valid object now so you cannot safely send messages to that object.
Then you do this:
NSLog(#”%i”,num);
In this NSLog statement, you are treating num as an integer, not as a pointer, because %i formats an integer. You're just printing the address (memory location) where the NSNumber object was (and might still be - you don't know). So it doesn't matter whether num points to a valid object or not; you're just treating num as an arbitrary int. You're not trying to send any messages to the NSNumber object.
By Calling [release] on object we can decrement retain count of the object.
If object is created by calling method which is having copy, new, alloc in it the caller is said to be owner of that object and owner
should alway release owned object.
If you are not owning the object you should not release it.
Always call release on object when object is retained which means if one call retain method the corresponding release call should be
invoked .Retain release should be matched.

Question about NSMutableArray, pointers and release

How exactly does the addObject method of NSMutableArray work? Does it create a new instance and add it into the array or does it simply add a reference to the SAME object into the array?
If the answer is it only insert a reference to the object, then it leads to my next question:
Let's say I have the following method in one of my class ('list' is a NSMutableArray), gladly, this code works the way I wanted, but i just don't seem to fully understand why:
-(void)buyItem:(Item *)anItem
{
Item * newItem = [[Item alloc]init];
newItem.name = anItem.name;
newItem.details = anItem.details;
[list addObject:newItem];
[newItem release];
}
So basically after calling [list addObject:newItem], there would now be total of two reference pointing to the same object right(newItem, and another one in the 'list' array)?
But why does releasing the newItem object here, doesn't wipe out the one in the 'list' NSMutableArray? Aren't they pointing to the same Object?
When you are adding object to NSMutableArray using method addObject: it retains added object. This is why you can release it later and use afterwards by accessing using objectAtIndex: method.
It adds a reference and then increases the objects retain count by one. What you are doing is correct and it will still exist in the array with a retain count of one.
For your reference.
What increases an object's retain count?
It's important to understand the distinction between release and dealloc. release simply decrements the "retain count", except that when the count is decremented to zero, release goes on to dealloc the object.
In general (except where documented otherwise), when you pass an object reference (ie, pointer) to an Objective-C object, and it keeps a copy of that reference beyond the duration of your call to it, it retains the object on its own behalf, and it takes the responsibility to release the object when it is itself deallocated, or when the copy of the reference is nullified or overwritten.

what's the difference between two code scenario

Scenario1:
NSDictionary *dictionary =
[[NSDictionary alloc] initWithContentsOfFile:plistPath];
self.stateZips = dictionary;
[dictionary release];
Scenario2:
self.stateZips = [[NSDictionary alloc] initWithContentsOfFile:plistPath];
dependes on stateZips property.
If it is retained:
Scenario 1: stateZips is properly retained ( a release on stateZips will call its dealloc). also local dictionary is released then and there.
Scenario 2: stateZips is retained twice ( a release in stateZips will not call its dealloc as it is still retained).
If it is assigned:
Scenario 1: stateZips points to released dictionary and accessing it else where might result in crash.
Scenario 2: stateZips is properly retained ( a release on stateZips will call its dealloc).
copy is not being considered, as i believe its not your intention (at least in this piece of code)
Both cause self.stateZips to be set to a dictionary initialized with the file pointed to in plistPath.
But in the second, the pointer to the initialized dictionary was not saved, and as it's an object with a retain count of +1 technically a release message needs to be sent to it in some place, to balance the memory management. But as there is no way to retrieve the pointer to that object, you'll end up with a memory leak.
Two exceptions apply:
1.Garbage Collection
If you're in a garbage collected environment, both are the same. Well, they are not the same, but the result is similar.
2.Property type
If the setter for stateZips simply assigns the pointer, then you can release the object using the ivar pointer. Then these two pieces of code have only one difference: in the former, the object is released right after it's used. In the latter, it's just "undefined". Without the context, it's hard to determine if this object was released or not, and when.
I am assuming that stateZips is a property with the retain attribute.
In Scenario 1. A dictionary is created with a retain count of 1 in the first line. In the second line the property will call retain again, increasing the retain count to 2. Finally the retain count is decremented by the release. This will leave the dictionary with the correct retain count.
In Scenario 2, the retain is only called once.
The net effect of the two scenarios is the same. The dictionary object will be retained, and you will need to include a release in the dealloc method of the class.
If this were not correctly handled by the compiler, it would be very hard indeed following the retain/release rules of objective-c.