Question

在Scala中是否可以编写如下内容：

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
}

或者我是否需要逐个实现每个Road的方法，并在roadway上调用相应的方法？

Answer 1

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

只要您不需要随身携带原始Bridge（例如，您要将Bridge存储在Road的集合中。< / p>

如果您确实需要再次返回Bridge，可以在owner中添加Road方法，该方法返回Any，使用构造函数参数设置它BridgeCauseway，然后进行模式匹配以获得您的桥梁：

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
  ...
}

Answer 2

如果，您可以Bridge trait进行排序。

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

如何委托Scala中的成员？

2 个答案: