Swift 快速弱自我功能保留
如果我有一个引用弱var weakSelf=self的闭包,我可以通过Swift 快速弱自我功能保留,swift,automatic-ref-counting,Swift,Automatic Ref Counting,如果我有一个引用弱var weakSelf=self的闭包,我可以通过weakSelf将该闭包更改为直接函数引用吗 struct ClosureHolder { let closure: () -> Void } class ClosureSource { func hello() { NSLog("hello") } func createWeakSelfWithinInnerClosureClosureHolder() -> C
weakSelf
将该闭包更改为直接函数引用吗
struct ClosureHolder {
let closure: () -> Void
}
class ClosureSource {
func hello() {
NSLog("hello")
}
func createWeakSelfWithinInnerClosureClosureHolder() -> ClosureHolder {
weak var weakSelf = self
return ClosureHolder(closure: {
weakSelf?.hello()
})
}
func createWeakSelfDotHelloClosureHolder() -> ClosureHolder {
weak var weakSelf = self
// The code below won't compile because weakSelf is an Optional.
// Once I unwrap the optional, I no longer have a weak reference.
// return ClosureHolder(closure: weakSelf.hello)
// this strongifies the weak reference. :(
return ClosureHolder(closure: weakSelf!.hello)
}
}
我更喜欢像
createWeakSelfDotHelloClosureHolder
,而不是createweakselfwithinnelclosureholder
,不,你不能。说self.foo
(如果foo
是一种方法)与说MyClass.foo(self)
完全一样。以这种方式出现的方法总是保持对接收者对象的强引用。如果您想要维护弱引用,那么您需要坚持使用{weakSelf?.hello()}
方法。您在这里试图做什么,而这些都不是由您处理的?我不知道闭包捕获列表。我认为这可能激发了一个更好的解决方案。