2016-08-24 72 views
3

我想知道爲什麼這個代碼中的akka​​ Actor的接收方法的行爲像一個val?爲什麼在akka中接收Actor方法的行爲像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() 

} 

輸出:

in receive 
Hello Akka 
Hello Akka 
Hello Akka 

,當它被認爲是:

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

而接收是DEF。

有關這種行爲的任何想法,爲什麼def這裏計算像val?

回答

4

receive返回一個PartialFunction [Any,Unit],並且該PartialFunction被播種爲Actors行爲(可以使用context.become/unbecome進行更改)。

它被認爲是val的原因是PartialFunction實例被重用,直到被更改。

+0

謝謝,但爲什麼在這[代碼](https://www.jdoodle.com/embed/v0/scala/2.12.0/14e)它的行爲像一個def? – SaKou

+0

您鏈接的代碼與PartialFunctions完全無關,您可以通過將返回類型更改爲Unit來證明這一點。您的方法在PartialFunction之外執行副作用(printlns)。 Akka調用(緩存的)PartialFunction。 –

+0

你可以請,看看上面的代碼,我已經將部分類型更改爲:type Partial = PartialFunction [Any,Unit] 我看不出差異 – SaKou

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