转义闭包是否捕获强链接？

Question

在我的示例中，我在其他函数中创建了实例。在函数结束时，我希望 instance2 应该是 nil 并且 completionHandlers 数组不应该对 SomeClass2 有强 link 但 completionHandlers 仍然 link on.

看起来 @escaping 闭包在 self 内部创建了强大的 links。

var completionHandlers: [() -> Void] = []

func someFunctionWithEscapingClosure(completionHandler: @escaping () -> Void) {
    completionHandlers.append(completionHandler)
}

class SomeClass {
    var x = 10
    func doSomething() {
        let instance2 = SomeClass2()
        instance2.doSomething2()
    }
}

class SomeClass2 {
    var x = 11
    func doSomething2() {
        someFunctionWithEscapingClosure {
            // still exist
            self.x = 77
        }
    }
}

let instance = SomeClass()
instance.doSomething()

completionHandlers.first!()

Answer 1

来自documentation：

If you assign a closure to a property of a class instance, and the closure captures that instance by referring to the instance or its members, you will create a strong reference cycle between the closure and the instance. Swift uses capture lists to break these strong reference cycles. For more information, see Strong Reference Cycles for Closures.

你这里没有强引用循环，但有人提到闭包捕获了实例，这就是你的情况。为防止您可以显式地捕获它：

func doSomething2() {
    someFunctionWithEscapingClosure { [weak self] in
        self?.x = 77
    }
}

转义闭包是否捕获强链接？

Escaped closure capture strong links or not?

closures

escaping

swift

swift3