我是Scala的新手,所以我需要帮助。我有一个用Java完成的示例,我必须使用Scala创建相同的东西。例子是这样的:

import java.util.ArrayList;

public class Z
{
    public static void main(String[] args)
    {
        A a=new A();
        B b=new B(a);
        a.setB(b);
        a.addMessage("A1");
        a.addMessage("A2");
        a.addMessage("A3");
        b.addMessage("B1");
        b.addMessage("B2");
        b.addMessage("B3");
        b.addMessage("B4");
        a.start();
        b.start();
    }
}

class A extends Thread
{
    ArrayList<String> msgs=new ArrayList<String>();
    B b;

    public void setB(B tb)
    {
        b=tb;
    }

    public void run() //It's like a send() method
    {
        for (int i=0; i<msgs.size(); i++)
        {
            final String msg=msgs.get(i);
            new Thread()
            {
                public void run()
                {
                    b.receive(msg);
                }
            }.start();
        }
    }

    public void receive (String msg)
    {
        System.out.println("A received a message from B.");
    }

    public void addMessage(String message)
    {
        msgs.add(message);
    }
}

class B extends Thread
{
    ArrayList<String> msgs=new ArrayList<String>();
    A a;

    public B(A ta)
    {
        a=ta;
    }

    public void run() //It's like a send() method
    {
        for (int i=0; i<msgs.size(); i++)
        {
            final String msg=msgs.get(i);
            new Thread()
            {
                public void run()
                {
                    a.receive(msg);
                }
            }.start();
        }
    }

    public void receive (String msg)
    {
        System.out.println("B received a message from A.");
    }

    public void addMessage(String message)
    {
        msgs.add(message);
    }
}


两个actor的创建是相似的,但是问题出在run()上。我知道我可以用这样的东西

receive
{
    case msg=>println("A received a message from B.");
}


在两个act()中接收消息(我必须说这非常有用),但是我不知道如何重新创建发送消息的部分。我应该从主角开始演员,然后在act()之外发送消息吗?

最佳答案

考虑以下与Akka参与者的实施。注意


run方法在这里被替换为SendAll消息;
通过将这些引用添加到SendAll可以简化演员之间传递的演员参考;
对代码简洁性的一个小改进是将每个文本消息封装到列表上;
actor日志记录系统替换println


因此,

import akka.actor.{ActorSystem, Props, Actor, ActorRef, ActorLogging}
import collection.mutable.ListBuffer

sealed trait Messaging
case class Add(msgs: List[String]) extends Messaging
case class Receive(msgs: String) extends Messaging
case class SendAll(dest: ActorRef) extends Messaging


class Messenger extends Actor with ActorLogging {
  var messages = new ListBuffer[String]()

  def receive = {
    case Add(msgs)     => messages ++= msgs
    case Receive(msg)  => log.info(s"received $msg from ${sender.path}.")
    case SendAll(dest) => messages.foreach(dest ! Receive(_))
  }
}

object Main extends App {

  val system = ActorSystem("MessengerSystem")
  val aActor = system.actorOf(Props[Messenger], name = "aActor")
  val bActor = system.actorOf(Props[Messenger], name = "bActor")

  aActor ! Add(List("A1","A2","A3"))
  bActor ! Add(List("B1","B2","B3","B4"))

  aActor ! SendAll(bActor)
  bActor ! SendAll(aActor)

  system.shutdown
}

10-07 20:39