In firstViewController,
SecondViewController *secondViewController = [[[SecondViewController alloc] init] autorelease];
// Pass data to secondViewController
[secondViewController setClass1:anObjectFromFirstViewController];
[[self navigationController] pushViewController:secondViewController ...];
SecondViewController.m
Class1 *class1;
NSMutableArray *object2;
In viewDidLoad, I think object2 points to the same object as class1 and therefore does not need to be allocated.
[self setObject2:[class1 someNSMutableArray]];
In init, I don't have to allocate object2 and the whole app still works. Is it because the object2 is pointing to the same item in memory as [cla开发者_C百科ss1 someNSMutableArray]
.
If that is the case, then what happens if I do allocate object2. Will there be 2 copies? The app still works the same if I initialize it.
-(id)init { object2 = [[NSMutableArray alloc] init]; }
The confusing part is that if I initialize with autorelease, it will give me a dealloc error, the deallocating to a released object error
-(id)init { object2 = [[[NSMutableArray alloc] init] autorelease]; }
Thanks mucho!!!
Your idea on the first part of the question is correct.
[self setObject2:[class1 someNSMutableArray]];
object2 is a variable which now contains the same address as [class1 someNSMutableArray]. So both point to the same NSMutableArray object.
This is different from:
object2 = [[NSMutableArray alloc] init];
In this case you have created a new NSMutableArray object and put that address in the object2 instance variable. So that is now pointing to a different array than [class1 someNSMutableArray].
The key word there is "created". Think of alloc/init as creating a new object.
As for memory management:
object2 = [[[NSMutableArray alloc] init] autorelease];
I'm assuming you are (correctly) releasing object2 in the -dealloc for this class. So you have two releases (that release in dealloc plus this autorelease here) against only one retain (from the alloc/init), and therefore you are over-releasing the object.
精彩评论