我如何委托给 Scala 中的成员？

发布于 2024-10-28 05:32:02 字数 340 浏览 3 评论 0原文

在Scala中是否可以写这样的东西：

trait Road {
  ...
}

class BridgeCauseway extends Road {
  // implements method in Road
}

class Bridge extends Road {
  val roadway = new BridgeCauseway()

  // delegate all Bridge methods to the `roadway` member
}

或者我需要一一实现Road的每个方法，并在roadway上调用相应的方法？

原文

Is it possible in Scala to write something like:

trait Road {
  ...
}

class BridgeCauseway extends Road {
  // implements method in Road
}

class Bridge extends Road {
  val roadway = new BridgeCauseway()

  // delegate all Bridge methods to the `roadway` member
}

or do I need to implement each of Road's methods, one by one, and call the corresponding method on roadway?

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。

鯉魚旗 2024-11-04 05:32:02

实现此目的的最简单方法是使用隐式转换而不是类扩展：

class Bridge { ... }
implicit def bridge2road(b: Bridge) = b.roadway

只要您不需要携带原始的 Bridge （例如，您要存储Road 集合中的 Bridge）。

如果您确实需要再次取回 Bridge，您可以在 Road 中添加 owner 方法，该方法返回 Any >，使用 BridgeCauseway 的构造函数参数进行设置，然后进行模式匹配以获取桥梁：

trait Road {
  def owner: Any
  ...
}

class BridgeCauseway(val owner: Bridge) extends Road { . . . }

class Bridge extends Road {
  val roadway = new BridgeCauseway(this)
  ...
}

myBridgeCauseway.owner match {
  case b: Bridge => // Do bridge-specific stuff
  ...
}

The easiest way to accomplish this is with an implicit conversion instead of a class extension:

class Bridge { ... }
implicit def bridge2road(b: Bridge) = b.roadway

as long as you don't need the original Bridge to be carried along for the ride (e.g. you're going to store Bridge in a collection of Road).

If you do need to get the Bridge back again, you can add an owner method in Road which returns an Any, set it using a constructor parameter for BridgeCauseway, and then pattern-match to get your bridge:

trait Road {
  def owner: Any
  ...
}

class BridgeCauseway(val owner: Bridge) extends Road { . . . }

class Bridge extends Road {
  val roadway = new BridgeCauseway(this)
  ...
}

myBridgeCauseway.owner match {
  case b: Bridge => // Do bridge-specific stuff
  ...
}

回复收藏 0 原文

爱殇璃 2024-11-04 05:32:02

如果您可以将Bridge作为特质，您就会被排序。

scala> trait A {
     |   val x: String
     | }
defined trait A

scala> class B extends A {
     |   val x = "foo"
     |   val y = "bar"
     | }
defined class B

scala> trait C extends A { self: B =>         
     |   val z = "baz"               
     | }
defined trait C

scala> new B with C
res51: B with C = $anon$1@1b4e829

scala> res51.x
res52: java.lang.String = foo

scala> res51.y
res53: java.lang.String = bar

scala> res51.z
res54: java.lang.String = baz

If you can make Bridge a trait you'll be sorted.

scala> trait A {
     |   val x: String
     | }
defined trait A

scala> class B extends A {
     |   val x = "foo"
     |   val y = "bar"
     | }
defined class B

scala> trait C extends A { self: B =>         
     |   val z = "baz"               
     | }
defined trait C

scala> new B with C
res51: B with C = $anon$1@1b4e829

scala> res51.x
res52: java.lang.String = foo

scala> res51.y
res53: java.lang.String = bar

scala> res51.z
res54: java.lang.String = baz

回复收藏 0 原文

~没有更多了~