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Let's say that I declare property in following way:

@property(nonatomic, strong, getter = isWorking) BOOL working;

Then instead of having the property to be synthesized I write the getter myself (and add some custom logic to it).

What will happen if I access the property in following way:

BOOL work = self.working;

Is the getter (and my custom logic there) still called or is it called only when I access the property using getter explicitly (BOOL work = self.isWorking;) ?

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1 回答 1

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哎呀。 刚试了一下。显然我使用点符号太多了，并没有意识到它做了多少。:P

#import "NSObject.h"
#include <stdio.h>

@interface Test : NSObject
@property (getter=myStuff) int stuff;
@end

@implementation Test
-(int)myStuff { return 42; }
-(void)setStuff:(int)value { /* don't care */ }
@end


int main() {
    @autoreleasepool {
        Test* test = [[Test alloc] init];

        /* All these work... */
        printf("test.stuff == %d\n", test.stuff);
        printf("[test myStuff] == %d\n", [test myStuff]);
        printf("test.myStuff == %d\n", test.myStuff);

        /* but here, there's an exception */
        printf("[test stuff] == %d\n", [test stuff]);

        return 0;
    }
}

当我编译这个（在 Linux 中使用 clang）时，有两个关于缺少的奇怪的警告-(int)stuff。输出看起来像

chao@chao-VirtualBox:~/code/objc$ ./a.out
test.stuff == 42
[test myStuff] == 42
test.myStuff == 42
: Uncaught exception NSInvalidArgumentException, reason: -[Test stuff]: unrecognized selector sent to instance 0x2367f38
chao@chao-VirtualBox:~/code/objc$

所以，嗯，是的。忽略下面一半的东西。:P

self.working只是语法糖[self working]（或者[self setWorking:value]如果你分配给它）。任何一个都会做同样的事情：返回 的值 [self isWorking]，因为那是您定义的 getter。

如果您想避免使用吸气剂，请尝试_working或self->_working（或您命名为 ivar 的任何名称）。否则，、self.working和（ 即使你感觉很勇敢）都应该给你相同的结果。[self working][self isWorking]self.isWorking

于 2013-08-17T17:55:07.587 回答