如何访问“覆盖"? Scala的内部阶级?

问题描述

我有两个特征，一个扩展了另一个，每个具有一个内部类，一个扩展了另一个，具有相同的名称:

I have two traits, one extending the other, each with an inner class, one extending the other, with the same names:

trait A {
    class X {
        def x() = doSomething()
    }
}

trait B extends A {
    class X extends super.X {
        override def x() = doSomethingElse()
    }
}

class C extends B {
    val x = new X() // here B.X is instantiated
    val y = new A.X() // does not compile
    val z = new A.this.X() // does not compile
}

如何访问C类主体中的A.X类?重命名B.X不隐藏A.X不是首选方法.

How do I access A.X class in the C class's body? Renaming B.X not to hide A.X is not a preferred way.

让事情有点复杂，在这种情况下，我遇到了这个问题，即特征具有类型参数(在此示例中未显示).

To make things a bit complicated, in the situation I have encountered this problem the traits have type parameters (not shown in this example).

推荐答案

trait A {
    class X {
        def x() = "A.X"
    }
}

trait B extends A {
    class X extends super.X {
        override def x() = "B.X"
    }
}

class C extends B {
  val self = this:A
  val x = new this.X()
  val y = new self.X()
}

scala> val c = new C
c: C = C@1ef4b

scala> c.x.x
res0: java.lang.String = B.X

scala> c.y.x
res1: java.lang.String = A.X

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