Re: In dealloc(): ref @property, Can I use "<property object> = nil; " vs "[<property object> release]; " ?
Re: In dealloc(): ref @property, Can I use "<property object> = nil; " vs "[<property object> release]; " ?
- Subject: Re: In dealloc(): ref @property, Can I use "<property object> = nil; " vs "[<property object> release]; " ?
- From: Negm-Awad Amin <email@hidden>
- Date: Wed, 8 Oct 2008 18:26:27 +0200
Am Mi,08.10.2008 um 18:09 schrieb Lee, Frederick:
I've been setting my property thusly:
@property (nonatomic, retain) myVar;
...what about:
@property (nonatomic, copy) myVar; ?
In this case (deallocation) it doesn't matter, because you set nil.
[nil copy] is nil.
Cheers
Ric.
From: Ron Lue-Sang [mailto:email@hidden]
Sent: Wednesday, October 08, 2008 11:06 AM
To: Lee, Frederick
Cc: email@hidden
Subject: Re: In dealloc(): ref @property, Can I use "<property
object> =
nil; " vs "[<property object> release]; " ?
On Oct 8, 2008, at 8:49 AM, Lee, Frederick wrote:
Assuming the following:
@property(retain) myVar;
...
@synthesize myVar;
...
-(void) dealloc {
// Can I use:
self.myVar = nil;
// versus:
[myVar release]; // ?
}
I've seen examples of using [myVar release]. But doesn't setting
myVar
= nil does the same thing?
Which is the preferred way?
I prefer doing [myVar release]. Mainly because I don't want setMyVar
getting called during dealloc. And by this time there *shouldn't* be
any
KVO observers for myVar (assuming the property was observed at all
before).
Yea yea - you've done @synthesize here so we know that the setter
doesn't do any custom work. But since we're in dealloc, we are -
obviously - not running under GC. Under refcounting, if you didn't
declare your property nonatomic, you're gonna take a lock during
dealloc
to do setMyVar:nil.
--------------------------
RONZILLA
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Amin Negm-Awad
email@hidden
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