Scala actor可以同时处理多条消息吗？

Question

回复到我最近的问题表明,一个演员处理其消息一次一个.这是真的？我没有看到任何明确说明(在Scala中编程),其中包含以下代码段(第593页)

如果[ react方法]找到可以处理的消息,[it]将调度该消息的处理以便以后执行并抛出异常

(强调我自己).两个相关(和互斥)的问题:

1. 假设演员可以同时处理多个消息,我怎样才能强制演员一次处理消息1(如果这是我想做的话)？(用receive？)
2. 假设一个actor一次处理一个消息,我怎样才能最好地实现一个实际上可以同时处理消息的actor

编辑:做一些测试似乎证明我错了,演员确实是连续的.所以问题#2我需要回答

Answer 1

Actor一次处理一条消息.处理多个消息的经典模式是为消费者角色池设置一个协调者角色.如果使用react,则使用者池可能很大,但仍然只使用少量JVM线程.这是一个例子,我为他们创建了一个由10个消费者和一个协调员组成的池.

import scala.actors.Actor
import scala.actors.Actor._

case class Request(sender : Actor, payload : String)
case class Ready(sender : Actor)
case class Result(result : String)
case object Stop

def consumer(n : Int) = actor {
  loop {
    react {
      case Ready(sender) => 
        sender ! Ready(self)
      case Request(sender, payload) =>
        println("request to consumer " + n + " with " + payload)
        // some silly computation so the process takes awhile
        val result = ((payload + payload + payload) map {case '0' => 'X'; case '1' => "-"; case c => c}).mkString
        sender ! Result(result)
        println("consumer " + n + " is done processing " + result )
      case Stop => exit
    }
  }
}

// a pool of 10 consumers
val consumers = for (n <- 0 to 10) yield consumer(n)

val coordinator = actor {
  loop {
     react {
        case msg @ Request(sender, payload) =>
           consumers foreach {_ ! Ready(self)}
           react {
              // send the request to the first available consumer
              case Ready(consumer) => consumer ! msg
           }
         case Stop => 
           consumers foreach {_ ! Stop} 
           exit
     }
  }
}

// a little test loop - note that it's not doing anything with the results or telling the coordinator to stop
for (i <- 0 to 1000) coordinator ! Request(self, i.toString)

此代码测试以查看哪些消费者可用并向该使用者发送请求.替代方案是随机分配给消费者或使用循环调度程序.

根据您的工作情况,您可能会更好地使用Scala的期货.例如,如果你真的不需要演员,那么所有上述机器都可以写成

import scala.actors.Futures._

def transform(payload : String) = {      
  val result = ((payload + payload + payload) map {case '0' => 'X'; case '1' => "-"; case c => c}).mkString
  println("transformed " + payload + " to " + result )
  result
}

val results = for (i <- 0 to 1000) yield future(transform(i.toString))

Answer 2

我认为答案是Actor无法异步处理消息.如果你有一个Actor应该监听消息,这些消息可以异步处理,那么它可以这样写:

val actor_ = actor {

  loop {
    react {
      case msg =>
        //create a new actor to execute the work. The framework can then 
        //manage the resources effectively
        actor {
          //do work here
        }
      }
    }
  }