Question

object Test1 extends App {

  object specificTypes extends Enumeration {
    type specificTypes = Value
    val Red = Value
    val Green = Value
    val Blue = Value
  }

  abstract class outer {
    type T <: Test1.specificTypes.Value
    def pr(w: T)
  }

  class red extends outer {
    type T = Test1.specificTypes.Red // Getting Error Here, type Red is not a member of object Test1.specificTypes
    def pr(w: T) = println("hello red")
  }

  val r = new red
  r.pr(Test1.specificTypes.Red)

}

类型Red不是对象Test1.specificTypes的成员，如何解决它，我要去哪里错误

Answer 1

我认为如果你想保持类似的结构，最好将类型T存储在伴侣对象中。否则你可以使用和类型并执行类似的操作

sealed trait Colour

trait Red extends Colour

trait Green extends Colour

class RedClass extends Red
class GreenClass extends Green


def matchMethod(check:Colour) = {        

    check match {

        case Red => println("hello this is red")
        case Green => println("hello this is green")
    }
}

通过这种方式，您可以完全匹配且安全地匹配，而无需奇怪的类型引用。

Answer 2

type T = Test1.specificTypes.Red没有分配类型。它尝试将值指定为类型，当然不会编译。

这个怎么样？

object Test1 extends App {


  object specificTypes extends Enumeration {
    type specificTypes = Value
    val Red = Value
    val Green = Value
    val Blue = Value
  }

  abstract class outer {
    type T <: Test1.specificTypes.Value
    def pr(w: T)
  }

  class red extends outer {
    type T = Test1.specificTypes.Red.type //IMPORTANT THING!
    def pr(w: T) = println("hello red")
  }
  class blue extends outer {
    type T = Test1.specificTypes.Blue.type //IMPORTANT THING!
    def pr(w: T) = println("hello blue")
  }

  val r = new red
  //r.pr(Test1.specificTypes.Blue) // compilation error
  /*
    Error:(23, 28) type mismatch;
   found   : Test1.specificTypes.Blue.type (with underlying type Test1.specificTypes.Value)
   required: Test1.r.T
      (which expands to)  Test1.specificTypes.Red.type
    r.pr(Test1.specificTypes.Blue)
   */


  val b = new blue
  b.pr(Test1.specificTypes.Blue) // compiles, prints hello blue
}

但是，对于你的情况（就像你提到的那样）

我有一些类型的类，我想区分模式匹配中的类别

我会使用一个简单的sealed trait，就像Chobeat的回答一样。

在scala中键入匹配

2 个答案: