在scala中定义时自引用val

时间:2015-11-29 04:27:03

标签: scala recursion functional-programming akka

这是我在scala中发现的并且它有效并且我无法弄清楚为什么,有人可以解释为什么这有效吗?

基本上我可以使用val的引用,而我仍在定义它(因为我的其他对象/ actor将其作为参数接受)

val backbone: ActorRef = context.actorOf(
  F_BackBone.props(
    context.actorOf(F_PictureHandler.props(backbone)), 
    context.actorOf(F_UserHandler.props(backbone)), 
    context.actorOf(F_PageProfileHandler.props(backbone))
  )
)

如果我没有明确定义类型,我将收到编译器错误,这是有道理的。

2 个答案:

答案 0 :(得分:2)

请注意,在这种特定情况下,虽然代码编译,但它不会正常运行,因为正如在另一个答案的评论中所建议的那样,传递给其他actor的backbone的值是' null& #39 ;.

此示例演示了这一点:

import akka.actor.{Props, Actor, ActorRef, ActorSystem}

class SenderReceiver(sendTo:ActorRef) extends Actor{

  override def preStart(): Unit = {
    self ! "Start"
  }

  def receive = {
    case "Start" => sendTo ! "Hello"
    case "Hello" => println("Received Hello")
  }
}

object SenderReceiver {
  def props(sendTo:ActorRef):Props = Props(new SenderReceiver(sendTo))
}

object Example extends App {

  val system = ActorSystem()

  val actor: ActorRef = system.actorOf(SenderReceiver.props(actor))
  system.awaitTermination()
}

这会产生以下结果(反复出现,因为主管策略试图重新启动actor):

[info] [ERROR] [12/01/2015 09:47:04.543] [default-akka.actor.default-dispatcher-9] [akka://default/user/$a] null
[info] java.lang.NullPointerException
[info]  at SenderReceiver$$anonfun$receive$1.applyOrElse(example.scala:10)
[info]  at akka.actor.Actor$class.aroundReceive(Actor.scala:467)
[info]  at SenderReceiver.aroundReceive(example.scala:3)
[info]  at akka.actor.ActorCell.receiveMessage(ActorCell.scala:516)
[info]  at akka.actor.ActorCell.invoke(ActorCell.scala:487)
[info]  at akka.dispatch.Mailbox.processMailbox(Mailbox.scala:238)
[info]  at akka.dispatch.Mailbox.run(Mailbox.scala:220)
[info]  at akka.dispatch.ForkJoinExecutorConfigurator$AkkaForkJoinTask.exec(AbstractDispatcher.scala:397)
[info]  at scala.concurrent.forkjoin.ForkJoinTask.doExec(ForkJoinTask.java:260)
[info]  at scala.concurrent.forkjoin.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:1339)
[info]  at scala.concurrent.forkjoin.ForkJoinPool.runWorker(ForkJoinPool.java:1979)
[info]  at scala.concurrent.forkjoin.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:107)

答案 1 :(得分:1)

在函数式编程语言中,递归定义是一个重要的概念。想想定义阶乘的经典例子。

对于Scala的具体案例,本月早些时候在一篇文章中给出了一个非常好的解释: A variable used in its own definition?