CODEWITHSUNDEEP
objective-cpointers
moshe
moshe

Reputation: 157

Objective-C pointer assignment and reassignment dilemma

If I do this:

1 NSMutableArray *near = [[NSMutableArray alloc] init];
2 NSMutableArray *all = [[NSMutableArray alloc] init];
3 NSMutableArray *current = near;
4 current = all;

What happens to near?

At line 3, am I setting current to point to the same address as near so that I now have two variables pointing to the same place in memory, or am I setting current to point to the location of near in memory such that I now have this structure:

current -> near -> NSMutableArray

The obvious difference would be the value of near at line 4. If the former is happening, near is untouched and still points to its initial place in memory. If the latter is happening,

Upvotes: 1

Views: 2071

Answers (4)

nacho4d
nacho4d

Reputation: 45158

current = all;

is just assigning all to current. Hence, you still have to release all when you are finished; [all release];. The same when assiging near to current; (Because you create them using alloc init)

Is exactly the behavior you get when you declare a property with assign attribute with the only difference you don't use self to access current: 

@property(nonatomic, assign) NSMutableArray *current;

you synthesize it:

@synthesize current;

and then assign:

NSMutableArray *near = [[NSMutableArray alloc] init];
NSMutableArray *all = [[NSMutableArray alloc] init];
...
self.current = all;  //in your program you will do: current = all;
self.current = near; //in your programm you will do: current = near;
...
[all release];
[near release];

in your dealloc method:

current = nil; //this is not necessary but good to do.
//see I am not releasing 'current'?

Alternatively, there is the retain approach (Explained by @Anders):

@property(atomic, retain) NSMutableArray *current;

self.current = all; //will be similar to: current = [all retain];

and in that case you will need to release current later

Upvotes: 0

AndersK
AndersK

Reputation: 36092

1 NSMutableArray *near = [[NSMutableArray alloc] init];

you have created an NSMutableObjectArray with retain count 1, 'near' points to it

2 NSMutableArray *all = [[NSMutableArray alloc] init];

you have created an NSMutableObjectArray with retain count 1, 'all' points to it

3 NSMutableArray *current = near;

'current' now points to the same object as 'near' does, i.e. the NSMutableArray with retain count 1

4 current = all;

'current' now points to the same object as 'all' does, i.e. the NSMutableArray with retain count 1

note the retain count, it is normally good when you reference an object to increase the retain count in order to be sure the object is still there if the other variable is released:

current = [all retain];
...
[current release];

Upvotes: 4

Matt Wilding
Matt Wilding

Reputation: 20163

You are setting current to point to the same location in memory that near is pointing to. In order for current to point to the memory location of near, it would have to say:

current = &near;

Upvotes: 1

omz
omz

Reputation: 53561

At line 3, current and near point to the same object (location in memory). You can simply log the memory addresses of both objects to validate this:

NSLog(@"current: %p / near: %p", current, near);

At line 4, near will still point to the same object.

Upvotes: 0

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