Question

在下面的代码片段中，ControllerActor向BackgroundActor请求多条消息。稍后询问的消息将在先前询问的消息之前处理。但有些如何不交付并导致DeadLetter异常。请在下面找到代码段： -

package pack

import scala.concurrent.duration._
import akka.actor.Actor
import akka.actor.ActorRef
import akka.actor.ActorSystem
import akka.actor.Props
import akka.pattern.ask
import akka.util.Timeout
import pack.ControllerActor.Msg

object ActorAskingMultipleMsgs {
  def main( args: Array[String] ): Unit = {
    val system = ActorSystem( "test-system" )
    val testActor = system.actorOf( Props[BackgroundActor], "testActor" )
    val controller = system.actorOf( ControllerActor.props( testActor ), "controller" )

    //second argument in Msg is the time
    //that is used in BackgroundActor to simulate delay
    controller ! Msg( "A", 8000 )
    controller ! Msg( "B", 6000 )
    controller ! Msg( "C", 1000 )

    Thread.sleep(Integer.MAX_VALUE)
  }
}

object ControllerActor {
  case class Msg( msg: String, a: Int )
  def props( testActor: ActorRef ) = Props( new ControllerActor( testActor ) )
}

class ControllerActor( val af: ActorRef ) extends Actor {
  import scala.concurrent.ExecutionContext.Implicits.global
  implicit val timeout = Timeout( 10 seconds )

  def receive = {
    case msg: Msg =>
      println( s"starting processing for msg $msg" )
      af ? msg map {
        case str: String =>
          println( s"data returned is $str" )
        case _ => println( "Invalid data is returned" )
      }
      println( s"finished processing for msg $msg" )

    case _ => println( "Unknown msg type in controller" )
  }
}

class BackgroundActor extends Actor {
  def receive = {
    case Msg( str, a ) =>
      println( s"processing of msg($str, $a) starts here" )
      Thread.sleep( a )
      println( s"processing of msg($str, $a) ends here" )
      sender ! str.toUpperCase()
    case _ =>
      println( "Unknown msg" )
      sender ! false
  }
}

请建议，我在这里做错了什么？

Answer 1

这是ControllerActor中的超时。将其设置为：

require(["dijit/registry", 'dojo/domReady!'], function(registry){ console.log(registry.byId('my-id')) });

原因是ControllerActor将立即发送所有三条消息。因此，将创造三个未来。由于actor一次处理一个消息，因此BackgroundActor至少需要8 + 6 = 14秒才能向第二条消息发送响应。这大于你设置超时的时间。

超时异常将终止ControllerActor。这就是你死信的原因。

请求多个混乱的DeadLetter异常

1 个答案: