Question

我想知道为什么这段代码中的akka Actor的receive方法表现得像val？

import akka.actor.{ ActorRef, ActorSystem, Props, Actor }
import scala.concurrent.duration._


// Define Actor Messages
case class WhoToGreet(who: String)


// Define Greeter Actor
class Greeter extends Actor {
  def receive = {
    println("in receive")
    receiveHandler
  }


  def receiveHandler: Receive = {
    case WhoToGreet(who) => println(s"Hello $who")
  }
}

object HelloAkkaScala extends App {

  // Create the 'hello akka' actor system
  val system = ActorSystem("Hello-Akka")

  // Create the 'greeter' actor
  val greeter = system.actorOf(Props[Greeter], "greeter")

  // Send WhoToGreet Message to actor
  greeter ! WhoToGreet("Akka")

  greeter ! WhoToGreet("Akka")

  greeter ! WhoToGreet("Akka")


  //shutdown actorsystem
  system.terminate()

}

输出：

in receive
Hello Akka
Hello Akka
Hello Akka

应该是：

in receive
Hello Akka
in receive
Hello Akka
in receive
Hello Akka

虽然接收是def。

关于这种行为的任何想法，为什么这里的def计算像val？

Answer 1

receive返回PartialFunction [Any，Unit]，并且PartialFunction被播种为Actors行为（可以使用context.become / unbecome更改）。

将其视为val的原因是PartialFunction实例在更改之前重复使用。

为什么在akka中接收Actor方法的行为类似于val？

1 个答案: