Question

我有两个特征，一个扩展另一个，每个都有一个内部类，一个扩展另一个，具有相同的名称：

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A.X班级的C班级？重命名B.X不隐藏A.X不是首选方式。

为了使事情有点复杂，在我遇到这个问题的情况下，特征具有类型参数（在这个例子中没有显示）。

Answer 1

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

Answer 2

对于那些对这个奇特的问题感兴趣的人，我发现它也可以作为函数的返回值。由于我的特征A和B具有类型参数，因此应该导致更简洁的代码：

trait A[T, U, V] {
    class X {
        def x() = "A.X"
    }

    def a = this:A[T, U, V]
}

trait B[T, U, V] extends A[T, U, V] {
    class X extends super.X {
        override def x() = "B.X"
    }
}

class C extends B[SomeClass, SomeOtherClass, ThirdOne] {
    val aVerbose = this:A[SomeClass, SomeOtherClass, ThirdOne] // works but is a bit ugly
    val aConcise = a
    val x = new this.X()
    val y = new aVerbose.X()
    val z = new aConcise.X()
}

scala> val c = new C()
c: C = C@1e852be

scala> c.x.x()
res2: java.lang.String = B.X

scala> c.y.x()
res3: java.lang.String = A.X

scala> c.z.x()
res4: java.lang.String = A.X

如何在Scala中访问“重写”内部类？

2 个答案: