保留循环Swift闭包

Question

我正在考虑使用此模式(或类似的模式)来观察纯Swift对象上的更改。

这段代码有没有内存泄漏？

class Person {

   var name: String {
       didSet{
           self.listen?(self)
       }
   }

   var phone: Int {
       didSet{
           self.listen?(self)
       }
   }

   var listen: ((_ person: Person) -> Void)?

   init(withName N: String, andPhone P: Int) {
       self.name = N
       self.phone = P
   }
}

 class HouseHold {

   var people = [Person]()

   let changes: (_ person: Person) -> Void = { person in
       print(person.name)
   }

   init(withPeople P: [Person]){
       self.people = P
       for person in people {
           person.listen = self.changes
       }
   }
}

let jon = Person(withName: "Jon", andPhone: 1232344567)


let thais = Person(withName: "Thais", andPhone: 1232344567)


let apartment = HouseHold(withPeople: [jon, thais])

jon.phone = 1232399999
jon.name = "Jon Smith"
jon.phone = 9999999999
thais.name = "Thais Smith"

另外，假设我有一个UIViewController，它以同样的方式侦听来自Person对象的更改，而不是HouseHold。会不会有内存泄漏？

此外，欢迎对该模式的评论。我喜欢使用泛型，这样我就不必为我想要观察的每个属性创建一个函数。或者像代码目前所做的那样传递整个对象。

注意:在游乐场编译

注2:尝试在不使用reactive cocoa的情况下迁移到更多虚拟机架构。

Answer 1

因为：

HouseHold has Person, HouseHold has changes, Person has HouseHold.changes. 

上面没有循环。

如果您的代码更改为：

person.listen = self

情况是这样的：

HouseHold has Person, Person has HouseHold. 

您可以通过检查deinit方法来测试泄漏：

class Person {

    //...

    deinit {
        print("Person deinited")
    }
}

class HouseHold {

    //...

    deinit {
        print("HouseHold deinited")
    }
}