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如何在父母和孩子之间发送自定义数据?

akka scala

0

blue-sky · 技术社区 · 6 年前

以下代码:

package playground

import akka.actor.{Actor, ActorRef, ActorSystem, Props, _}
import akka.stream.ActorMaterializer

case object ReplyMessage
case object StopMessage
case object LinRMessage
case object LinRMessageChild

case class ListData(data : List[String])

class ModelParent(child: ActorRef) extends Actor {
  val logger = com.typesafe.scalalogging.Logger("ModelParent")

  def receive = {

    case ListData =>
      logger.info("Got reply in Parent")
    case LinRMessage =>
      child ! LinRMessageChild
    case _ => println("Parent got something unexpected.")
  }
}

class ModelChild extends Actor {
  val logger = com.typesafe.scalalogging.Logger("ModelChild")

  def receive = {
    case LinRMessageChild =>
      logger.info("Received LinRMessage")
      //sender ! ReplyMessage
      val l = List("1" , "2" , "3")
      sender ! ListData(l)
    case StopMessage =>
      println("Received Stop Message")
      context.stop(self)
    case _ => println("Child got something unexpected.")
  }
}

object ModelsDriver {

  def main(args: Array[String]): Unit = {

    val system = ActorSystem("sys")
    implicit val materializer = ActorMaterializer.create(system)

    val modelChild = system.actorOf(Props[ModelChild], name = "modelChild")
    val modelParent = system.actorOf(Props(new ModelParent(modelChild)), name = "modelParent")

    modelParent ! LinRMessage
    modelParent ! LinRMessage
    modelParent ! LinRMessage

  }

}

打印到控制台:

22:08:55.807 [sys-akka.actor.default-dispatcher-2] INFO ModelChild - Received LinRMessage
22:08:55.810 [sys-akka.actor.default-dispatcher-2] INFO ModelChild - Received LinRMessage
22:08:55.810 [sys-akka.actor.default-dispatcher-2] INFO ModelChild - Received LinRMessage
Parent got something unexpected.
Parent got something unexpected.
Parent got something unexpected.

下面是导致 Parent got something unexpected. 输出:

case _ => println("Parent got something unexpected.")

我试图将字符串列表从子级返回到类型为的父级参与者:

case class ListData(data : List[String])

这将填充返回数据:

  val l = List("1" , "2" , "3")
  sender ! ListData(l)

但错误表明我没有正确发送数据。如何在父参与者和子参与者之间发送消息中的数据?

1 回复 | 直到 6 年前

1

4

Jack Leow 6 年前

问题是由于你的父母演员的台词:

case ListData =>
  logger.info("Got reply in Parent")

通常,您希望模式匹配并提取case类的内容,如下所示:

case ListData(l: List[String]) =>
  logger.info("Got reply in Parent")

在这里,你可以像 l .

因为在您的示例中,您并没有对内容进行任何处理,所以您也可以执行以下操作之一:

case ListData(_) =>
  logger.info("Got reply in Parent")

或者:

case _: ListData =>
  logger.info("Got reply in Parent")

2

3

Mario Galic 6 年前

当我们定义一个case类时

case class Foo(x: Int)

编译器还自动定义相应的伴生单例对象

object Foo {
  def apply(x: Int): Foo = new Foo(x)
  def unapply(v: Foo): Option[Foo] = ...
}

这就是功能

val recieve: Any => String = {
  case Foo    => "I am companion singleton object of Foo"
  case Foo(_) => "I am instance of Foo case class"
}

recieve(Foo)
recieve(Foo(42))

输出

res0: String = I am companion singleton object of Foo
res1: String = I am instance of Foo case class

我们看到的地方

case Foo => "I am companion singleton object of Foo"

匹配自动生成的伴生单例对象。

同样地,

val recieve: Any => String = {
  case _: Foo.type => "I am companion singleton object of Foo"
  case _: Foo      => "I am instance of Foo case class"
}

recieve(Foo)
recieve(Foo(42))

输出

res0:String=我是Foo的同伴单例对象
res1:String=我是Foo case类的实例

因为它们是单例对象的类型 Foo 是 Foo.type ,而实例的类型 Foo(42) 是 福 也考虑

val a: Foo = Foo(42)
val b: Foo.type = Foo

这个逻辑错误是由于 Any 在里面 Any => Unit 另一方面 Foo => Unit 将在编译时捕获错误

val recieve: Foo => Unit = {
  case Foo =>  // compiler error: pattern type is incompatible with expected type
}