Swift：为什么我不能从override init调用方法？

发布时间：2020-12-14 02:29:33 所属栏目：百科来源：网络整理

导读：我有以下代码示例(来自PlayGround)： class Serializable : NSObject{override init() { }}class Device : Serializable{ var uuid:String override init() { println("init ") self.uuid = "XXX" self.uuid = Device.createUUID() println(self.uuid) } cla

我有以下代码示例(来自PlayGround)：

class Serializable : NSObject{
override init() { }
}

class Device : Serializable{

    var uuid:String

    override init() {

        println("init ")

        self.uuid = "XXX"

       self.uuid = Device.createUUID()

        println(self.uuid)

    }

   class func createUUID() -> String{
       return "XXX2"
    }
}


var device = Device()

您可以注意到我将createUUID方法实现为静态.

但是为什么我不能以静态的方式从init调用这个方法呢？：

class Serializable : NSObject{
override init() { }
}

class Device : Serializable{

    var uuid:String

    override init() {

        // tried
        // super.init()

        println("init ")

        self.uuid = "XXX"

       self.uuid = self.createUUID() //  ERROR
       self.uuid = createUUID() // ERROR

        println(self.uuid)

        // tried
        // super.init()

    }

  func createUUID() -> String{
       return "XXX2"
    }
}


var device = Device()

没有继承它可以正常工作：

class Device {

    var uuid:String

     init() {

        println("init ")

        self.uuid = "XXX"

       self.uuid = self.createUUID()

        println(self.uuid)

    }

    func createUUID() -> String{
       return "XXX2"
    }
}


var device = Device()

有两个竞争 initialization safety checks导致您的问题.

Safety check 1

A designated initializer must ensure that all of the properties introduced by its class are initialized before it delegates up to a superclass initializer.

和

Safety check 4

An initializer cannot call any instance methods,read the values of any instance properties,or refer to self as a value until after the first phase of initialization is complete.

下面是它的工作原理.

override init() {
    super.init() // Fails safety check 1: uuid is not initialized.
    uuid = createUUID()
}

反过来,

override init() {
    uuid = createUUID() // Fails safety check 4: cannot call an instance method before initialization is complete.
    super.init()
}

感谢@Ruben in his answer

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!