你是否模糊了商业Java代码？(Do you obfuscate your commercial Java code? [closed])

 在Objective-C中，属性中的copy属性意味着合成的setter将如下所示： 
 
-(void)setMasterBirdSightingList:(NSMutableArray*)newValue
{
    if (_masterBirdSightingList == newValue) return;
//  NSMutableArray* oldValue = _masterBirdSightingList;
    _masterBirdSightingList = [newValue copy];
//  [oldValue release];    // <-- not applicable in ARC.
}
 
 并且点语法将始终转换为 
 
[self setMasterBirdSightingList:sightingList];
 
 无论属性的属性如何。 
 
 “分配给另一个空间以保存来自sightingList的内容”的东西是通过-copy方法完成的。 将参数传递给setter的newValue参数的方式无关紧要。 
 
 
 编辑 ：正如注释中提到的@David一样 ，可变类型的-copy方法返回一个不可变对象。 您必须覆盖setter才能调用-mutableCopy 。 请参阅使用可变对象的Obj-C 2.0属性的最佳方法是什么，例如NSMutableArray？ 。 

In Objective-C, the copy attribute in a property means the setter synthesized will look like this:
 
-(void)setMasterBirdSightingList:(NSMutableArray*)newValue
{
    if (_masterBirdSightingList == newValue) return;
//  NSMutableArray* oldValue = _masterBirdSightingList;
    _masterBirdSightingList = [newValue copy];
//  [oldValue release];    // <-- not applicable in ARC.
}
 
and that dot syntax will always be translated to
 
[self setMasterBirdSightingList:sightingList];
 
regardless of the attribute of the property.
 
The "allocated for another space to preserve stuffs from sightingList" stuff is done via the -copy method. The way you pass the argument to the setter's newValue parameter is irrelevant.
 
 
Edit: As @David mentioned in the comment, the -copy method of a mutable type returns an immutable object. You have to override the setter to call -mutableCopy instead. See What's the best way to use Obj-C 2.0 Properties with mutable objects, such as NSMutableArray?.