实例化不可变的配对对象

发布于 2024-12-06 00:57:19 字数 380 浏览 0 评论 0原文

是否可以创建一个具有对合作伙伴对象的不可变引用的类，或者它是否必须是我在创建后分配的 var ？

例如

class PairedObject (p: PairedObject, id: String) {
  val partner: PairedObject = p  // but I need ref to this object to create p!
}

或类似地，我如何实例化以下对？

class Chicken (e: Egg) { 
  val offspring = e
}

class Egg (c: Chicken) {
  val mother = c
}

原文

Is it possible to create a class with an immutable reference to a partner object, or does it have to be a var that I assign after creation?

e.g.

class PairedObject (p: PairedObject, id: String) {
  val partner: PairedObject = p  // but I need ref to this object to create p!
}

or similarly how could I instantiate the following pair?

class Chicken (e: Egg) { 
  val offspring = e
}

class Egg (c: Chicken) {
  val mother = c
}

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

野稚 2024-12-13 00:57:19

这是鸡/蛋问题的完整解决方案：

class Chicken (e: =>Egg) { 
  lazy val offspring = e 
}

class Egg (c: =>Chicken) {
  lazy val mother = c
}

lazy val chicken: Chicken = new Chicken(egg)
lazy val egg: Egg         = new Egg(chicken)

请注意，您必须为chicken 和egg 变量提供显式类型。

对于配对对象：

class PairedObject (p: => PairedObject, val id: String) {
  lazy val partner: PairedObject = p
}

lazy val p1: PairedObject = new PairedObject(p2, "P1")
lazy val p2: PairedObject = new PairedObject(p1, "P2")

Here is a complete solution to the Chicken/Egg problem:

class Chicken (e: =>Egg) { 
  lazy val offspring = e 
}

class Egg (c: =>Chicken) {
  lazy val mother = c
}

lazy val chicken: Chicken = new Chicken(egg)
lazy val egg: Egg         = new Egg(chicken)

Note that you have to provide explicit types to the chicken and egg variables.

And for PairedObject:

class PairedObject (p: => PairedObject, val id: String) {
  lazy val partner: PairedObject = p
}

lazy val p1: PairedObject = new PairedObject(p2, "P1")
lazy val p2: PairedObject = new PairedObject(p1, "P2")

回复收藏 0 原文