如何在 Scala 中实例化由类型参数表示的类型实例 [英] How to instantiate an instance of type represented by type parameter in Scala

问题描述

示例:

import scala.actors._  
import Actor._  

class BalanceActor[T <: Actor] extends Actor {  
  val workers: Int = 10  

  private lazy val actors = new Array[T](workers)  

  override def start() = {  
    for (i <- 0 to (workers - 1)) {  
      // error below: classtype required but T found  
      actors(i) = new T  
      actors(i).start  
    }  
    super.start()  
  }  
  // error below:  method mailboxSize cannot be accessed in T
  def workerMailboxSizes: List[Int] = (actors map (_.mailboxSize)).toList  
.  
.  
.  

注意第二个错误表明它知道actor项是T"，但不知道T"是actor的子类，如类泛型定义中的约束.

Note the second error shows that it knows the actor items are "T"s, but not that the "T" is a subclass of actor, as constrained in the class generic definition.

如何更正此代码以使其正常工作(使用 Scala 2.8)?

How can this code be corrected to work (using Scala 2.8)?

推荐答案

EDIT - 抱歉，我才注意到你的第一个错误.无法在运行时实例化 T，因为编译程序时类型信息会丢失(通过类型 擦除)

EDIT - apologies, I only just noticed your first error. There is no way of instantiating a T at runtime because the type information is lost when your program is compiled (via type erasure)

你必须通过一些工厂来实现构建:

You will have to pass in some factory to achieve the construction:

class BalanceActor[T <: Actor](val fac: () => T) extends Actor {
  val workers: Int = 10

  private lazy val actors = new Array[T](workers)

  override def start() = {
    for (i <- 0 to (workers - 1)) {
      actors(i) = fac() //use the factory method to instantiate a T
      actors(i).start
    }
    super.start()
  }
} 

这可能与某些演员 CalcActor 一起使用，如下所示:

This might be used with some actor CalcActor as follows:

val ba = new BalanceActor[CalcActor]( { () => new CalcActor } )
ba.start

顺便说一句:您可以使用 until 而不是 to:

As an aside: you can use until instead of to:

val size = 10
0 until size //is equivalent to:
0 to (size -1)

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