使用依赖于路径的类型作为类参数 [英] Use of path-dependent type as a class parameter

问题描述

我想要一个类，它接受一个具有类相关类型的参数，就像我经常为方法所做的那样.然而，令我惊讶的是，这不起作用:

I want to have a class which takes a parameter with a class-dependent type, as I've often done for methods. However, to my surprise, this didn't work:

scala> trait Compiler { trait Config }
defined trait Compiler

// works fine, as expected
scala> def f(c: Compiler)(conf: c.Config) = {}
f: (c: Compiler)(conf: c.Config)Unit

scala> class F(val c: Compiler)(val conf: c.Config)
<console>:8: error: not found: value c
       class F(val c: Compiler)(val conf: c.Config)
                                          ^

为什么?有什么解决方法吗?

Why? And are there any workarounds?

推荐答案

一种看起来可以接受的解决方法(不能在没有额外强制转换的情况下创建无效的 F):

A workaround which seems acceptable (can't create an invalid F without additional casts):

class F private (val c: Compiler)(_conf: Compiler#Config) {
  def conf = _conf.asInstanceOf[c.Config]
}

object F {
  def apply(c: Compiler)(conf: c.Config) = new F(c)(conf)
}

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