一个难题：弱引用在 Swift 4 中不起作用？

Question

我在 Playground 中创建了以下两个类，并希望它们在超出范围后都可以解除分配（因为引用很弱）。但令我惊讶的是他们没有！有人可以阐明这个难题吗？提前谢谢你。

代码如下：

class User {
    var name: String
    weak var phone: Phone?    
    init(name: String) {
        self.name = name
        print("User \(name) is initialized")
    }    
    deinit {
        print("User \(name) is deallocated")
    }
}

class Phone {
    let model: String
    weak var user: User?
    init(model: String) {
        self.model = model
        print("Phone \(model) is initialized")
    }
    deinit {
        print("Phone \(model) is deallocated")
    }
}

do {
    let user1 = User(name: "John")
    let phone1 = Phone(model: "iPhone7")
    user1.phone = phone1
    phone1.user = user1
}

print("Done")

这是显示未调用 deinit() 的输出，尽管类超出范围：

User John is initialized
Phone iPhone7 is initialized
Done

这是为什么？

Answer 1

我认为这与 Playgrounds 处理代码的方式有关。它没有完全相同的生命周期，您不能期望在代码完成后释放变量运行。

如果您使对象成为可选对象，一切都会按预期解除分配：

do {
  let user1: User? = User(name: "John")
  let phone1: Phone? = Phone(model: "iPhone7")
  user1?.phone = phone1
  phone1?.user = user1
}

User John is initialized
Phone iPhone7 is initialized
Phone iPhone7 is deallocated
User John is deallocated
Done

一个难题：弱引用在 Swift 4 中不起作用？

A puzzle: weak reference not working in Swift 4?

swift

swift-playground