Objective-C Memory Issue

Question

I'm having an issue with the memory management in my application. I have an NSDictionary instance variable that I'm setting equal to another NSDictionary that gets made in a method. This all works fine and my application behaves like I want it to, but I'm having trouble applying the proper memory management.

If I release the local dictionary it eventually causes a crash as the method is called repeatedly, because the data saved in the instance variable is also trashed. Here's the code:

NSBundle *bundle = [NSBundle mainBundle];
NSString *plistPath = [bundle pathForResource:@"Names" ofType:@"plist"];

NSDictionary *dictionary = [[NSDictionary alloc] initWithContentsOfFile:plistPath];
self.dictAllValues = dictionary;
[dictionary release];

Chandan Shetty SP · Accepted Answer · 2011-02-26 18:47:16Z

2

Create dictAllValues using

 @property(retain) NSDictionary *dictAllValues;

Your method

-(void) myMethod
{
NSBundle *bundle = [NSBundle mainBundle];
NSString *plistPath = [bundle pathForResource:@"Names" ofType:@"plist"];

NSDictionary *dictionary = [[NSDictionary alloc] initWithContentsOfFile:plistPath];
self.dictAllValues = dictionary;
[dictionary release];
}

and release in dealloc method

-(void) dealloc
{
[dictAllValues release];
[super dealloc];
}

answered Feb 26, 2011 at 18:47

Chandan Shetty SP

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1 Comment

Chandan Shetty SP Over a year ago

P.S. If you are modifying the dictionary... use NSMutableDictionary instead of NSDictionary

bbum · Accepted Answer · 2011-02-26 18:22:33Z

1

How do you declare dictAllValues? Typically, it would be:

 @property(retain) NSDictionary *dictAllValues;

If so, then the release in your code is correct and your problem lies elsewhere. Post the backtrace of the crash, use Build and Analyze and fix any issues, and try turning on Zombie detection.

answered Feb 26, 2011 at 18:22

bbum

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Comments

bbum · Accepted Answer · 2011-02-27 18:08:32Z

1

From the apple memory management guide.

As a corollary of the fundamental rule, if you need to store a received object as a property in an instance variable, you must retain or copy it.

So, in this case putting [dictionary release]; in dealloc method instead (or any other method you might use for clean up) should work fine.

I assume your dictAllValues property uses simple assignment, let me know if that's not the case.

edited Feb 27, 2011 at 18:08

bbum

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answered Feb 26, 2011 at 18:04

Nikita Rybak

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7 Comments

Justin K Over a year ago

thanks for the quick reply, that explains a lot, but won't dictionary be out of scope in the dealloc method? yes, it uses simple assignment. is it dictAllValues that must be released in the dealloc method?

Nikita Rybak Over a year ago

@Justin What do you mean, "out of scope"? By creating an object through alloc you became an owner of an object and thus need to call release or autorelease when you're done. Right now you're signaling "I don't need this object anymore" in the current method and thus object is trashed (with its contents). If you plan to use it later (and you seem to do, as you've created a local property), save it for now.

Justin K Over a year ago

yeah, I know, that's what I'm trying to do, but dictionary is declared in a class method as a local variable; if I reference it in -(void)dealloc it's not going to know what I'm talking about, did you mean to put [dictAllValues release] in the dealloc method? I already have that in there, so I assume that it'd be fine the way it is?

Nikita Rybak Over a year ago

@Justin If the dictAllValues property uses simple assignment, then it references the same object as the local variable dictionary. Note, you don't release references, you release objects they point to. So, you create an object in the current method (init ?) and release it in dealloc. dictAllValues reference is used to locate object.

bbum Over a year ago

The code as OP posted would normally be correct, assuming that OP followed the traditional @property declaration for allDictValues. There isn't enough information to draw any kind of conclusion as this answer (and comments) have done.

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