参考Scala中内部类的类型 [英] Referring to the type of an inner class in Scala

问题描述

以下代码试图模仿 DSL 的多态嵌入:它不是在 Inner 中给出行为，而是在其封闭类的 useInner 方法中编码.我添加了 enclosure 方法，这样用户只需保留对 Inner 实例的引用，但始终可以获取它们的封闭实例.通过这样做，来自特定 Outer 实例的所有 Inner 实例都只绑定到一个行为(但这里需要它).

The following code tries to mimic Polymorphic Embedding of DSLs: rather than giving the behavior in Inner, it is encoded in the useInner method of its enclosing class. I added the enclosing method so that user has only to keep a reference to Inner instances, but can always get their enclosing instance. By doing this, all Inner instances from a specific Outer instance are bound to only one behavior (but it is wanted here).

abstract class Outer {
  sealed class Inner {
    def enclosing = Outer.this
  }
 def useInner(x:Inner) : Boolean
}

def toBoolean(x:Outer#Inner) : Boolean = x.enclosing.useInner(x)

它不能编译并且 scala 2.8 抱怨:

It does not compile and scala 2.8 complains about:

type mismatch; found: sandbox.Outer#Inner
               required: _81.Inner where val _81:sandbox.Outer

来自编程Scala:嵌套类和Scala 之旅:内部类，在我看来问题在于 useInner期望来自特定 Outer 实例的 Inner 实例作为参数.

From Programming Scala: Nested classes and A Tour of Scala: Inner Classes, it seems to me that the problem is that useInnerexpects as argument an Inner instance from a specific Outer instance.

真正的解释是什么以及如何解决这个问题?

What is the true explanation and how to solve this problem ?

推荐答案

我想 Inner 类型就像 this.Inner 类型.Outer#Inner 独立于外部实例(不是路径依赖类型).

I suppose the type Inner is like the type this.Inner. Outer#Inner is independent of the outer instance (not a path-dependent type).

abstract class Outer {
  sealed class Inner {
    def enclosing = Outer.this
  }
  def useInner(x:Outer#Inner) : Boolean
}

def toBoolean(x:Outer#Inner) : Boolean = x.enclosing.useInner(x)

这篇关于参考Scala中内部类的类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！