我得到ARC编译器的以下警告:

"performSelector may cause a leak because its selector is unknown".

这是我正在做的:

[_controller performSelector:NSSelectorFromString(@"someMethod")];

为什么我得到这个警告?我理解编译器不能检查选择器是否存在,但为什么会导致泄漏?我怎样才能改变我的代码,让我不再得到这个警告呢?


当前回答

Matt Galloway的回答解释了原因:

考虑以下几点: id anotherObject1 = [someObject performSelector:@selector(copy)]; id anotherObject2 = [someObject performSelector:@selector(giveMeAnotherNonRetainedObject)]; 现在,ARC如何知道第一个返回一个保留计数为1的对象,而第二个 返回一个自动释放的对象?

如果您忽略了返回值,那么抑制警告似乎通常是安全的。如果你真的需要从performSelector获取一个保留对象,我不确定最好的做法是什么——除了“不要这样做”。

其他回答

我的猜测是这样的:因为选择器对编译器是未知的,ARC不能强制执行适当的内存管理。

事实上,有时内存管理通过特定的约定与方法的名称绑定在一起。具体来说,我在考虑方便构造函数和make方法;前者按照约定返回一个自动释放的对象;后者是一个保留对象。该约定基于选择器的名称,因此如果编译器不知道选择器,那么它就不能强制执行正确的内存管理规则。

如果这是正确的,我认为你可以安全地使用你的代码,只要你确保内存管理一切正常(例如,你的方法不返回它们分配的对象)。

这里有很多答案,但由于这个有点不同,我把几个答案结合起来,我想我应该把它放进去。我使用一个NSObject类别检查,以确保选择器返回void,也抑制编译器警告。

#import <Foundation/Foundation.h>
#import <objc/runtime.h>
#import "Debug.h" // not given; just an assert

@interface NSObject (Extras)

// Enforce the rule that the selector used must return void.
- (void) performVoidReturnSelector:(SEL)aSelector withObject:(id)object;
- (void) performVoidReturnSelector:(SEL)aSelector;

@end

@implementation NSObject (Extras)

// Apparently the reason the regular performSelect gives a compile time warning is that the system doesn't know the return type. I'm going to (a) make sure that the return type is void, and (b) disable this warning
// See http://stackoverflow.com/questions/7017281/performselector-may-cause-a-leak-because-its-selector-is-unknown

- (void) checkSelector:(SEL)aSelector {
    // See http://stackoverflow.com/questions/14602854/objective-c-is-there-a-way-to-check-a-selector-return-value
    Method m = class_getInstanceMethod([self class], aSelector);
    char type[128];
    method_getReturnType(m, type, sizeof(type));

    NSString *message = [[NSString alloc] initWithFormat:@"NSObject+Extras.performVoidReturnSelector: %@.%@ selector (type: %s)", [self class], NSStringFromSelector(aSelector), type];
    NSLog(@"%@", message);

    if (type[0] != 'v') {
        message = [[NSString alloc] initWithFormat:@"%@ was not void", message];
        [Debug assertTrue:FALSE withMessage:message];
    }
}

- (void) performVoidReturnSelector:(SEL)aSelector withObject:(id)object {
    [self checkSelector:aSelector];

#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Warc-performSelector-leaks"
    // Since the selector (aSelector) is returning void, it doesn't make sense to try to obtain the return result of performSelector. In fact, if we do, it crashes the app.
    [self performSelector: aSelector withObject: object];
#pragma clang diagnostic pop    
}

- (void) performVoidReturnSelector:(SEL)aSelector {
    [self checkSelector:aSelector];

#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Warc-performSelector-leaks"
    [self performSelector: aSelector];
#pragma clang diagnostic pop
}

@end

如果你不需要传递任何参数,一个简单的解决方法是使用valueForKeyPath。这甚至可以在Class对象上实现。

NSString *colorName = @"brightPinkColor";
id uicolor = [UIColor class];
if ([uicolor respondsToSelector:NSSelectorFromString(colorName)]){
    UIColor *brightPink = [uicolor valueForKeyPath:colorName];
    ...
}

奇怪但事实是:如果可以接受(即result为void并且你不介意让runloop循环一次),添加一个延迟,即使这是零:

[_controller performSelector:NSSelectorFromString(@"someMethod")
    withObject:nil
    afterDelay:0];

这删除了警告,大概是因为它向编译器保证没有对象可以返回,并且在某种程度上管理不当。

Matt Galloway的回答解释了原因:

考虑以下几点: id anotherObject1 = [someObject performSelector:@selector(copy)]; id anotherObject2 = [someObject performSelector:@selector(giveMeAnotherNonRetainedObject)]; 现在,ARC如何知道第一个返回一个保留计数为1的对象,而第二个 返回一个自动释放的对象?

如果您忽略了返回值,那么抑制警告似乎通常是安全的。如果你真的需要从performSelector获取一个保留对象,我不确定最好的做法是什么——除了“不要这样做”。