How to instantiate an instance of type represented by type parameter in Scala

Question

example:

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   .   .   .   

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.

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

Answer 1

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)

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()   } }  

This might be used with some actor CalcActor as follows:

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

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

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