Question

我有一个属性

//.h
@property (nonatomic, retain) SomeCLass * someSynthInstance;
//.m
@synthesize someSynthInstance = _someSynthInstance;

- (void) foo 
{
    self.someSynthInstance = [[SomeCLass alloc] init];
    //[self.someSynthInstance release]; <- SHOULD I RELEASE HERE???
}

- (void) dealloc 
{
    self.someSynthInstance = nil;
}

我的理论认为，[alloc，init]创建一个计数1，并且setter，inc计数，因此它变为2，因此我应该在它之后立即释放它

但是我改变了这样的一切之后，目标是在应用程序中获取exc_bad_access，所以目标不确定它是否正常

Answer 1

您希望释放实例变量而不是属性。所以你可以这样做：

self.someSynthInstance = [[[SomeCLass alloc] init] autorelease]; // puts it in the autoreleasepool so it'll get released automatically at some point in the near future

或

_someSynthInstance = [[SomeCLass alloc] init]; // skip the property

或

self.someSynthInstance = [[SomeCLass alloc] init];
[_someSynthInstance release]; // call release on the instance variable

Answer 2

标准做法是：

SomeClass *tmpObj = [[SomeClass alloc] init];
self.someSynthInstance = tmpObj;
[tmpObj release];

Answer 3

你应该发布类变量，而不是像[_someSynthInstance release];那样应该做的。

Answer 4

是的，你应该在init之后发布

SomeCLass *temp = [[SomeCLass alloc] init];
self.someSynthInstance = temp;
[temp release]

在alloc init之后释放合成属性

4 个答案: