作为当前 actor 接收函数名

Question

我有一个具有 2 个行为的简单演员

package com.hello

import akka.actor.{Actor, ActorLogging}

case object Ping

class Hello extends Actor with ActorLogging {

  import context._

  def receive: Receive = behaviorFoo

  self ! Ping

  def behaviorFoo: Receive = {
    case _ =>
      log.info(this.receive.getClass.getSimpleName)
      become(behaviorBar)
      self ! Ping
  }

  def behaviorBar: Receive = {
    case _ =>
      log.info(this.receive.getClass.getSimpleName)
  }
}

这件事做了那些：

ping 自己

记录当前接收函数

将行为更改为 behaviorBar

ping 自己

记录当前接收函数

在这两种情况下，它都会记录“$anonfun$behaviorFoo$1”

为什么第二个日志中不是“$anonfun$behaviorBar$1”？

如果我将代码更改为

  self ! Ping

  def receive: Receive = {
    case _ =>
      log.info(this.receive.getClass.getSimpleName)
      become(behaviorFoo)
      self ! Ping
  }

  def behaviorFoo: Receive = {
    case _ =>
      log.info(this.receive.getClass.getSimpleName)
      become(behaviorBar)
      self ! Ping
  }

  def behaviorBar: Receive = {
    case _ =>
      log.info(this.receive.getClass.getSimpleName)
  }

它记录了 3 次“$anonfun$receive$1”

是否存在获取当前行为（接收）函数名称的方法？或者我需要随时像 log.info("behaviorFoo") 那样硬写？

更新：

对于日志记录问题，我添加了

trait MyLogging extends ActorLogging {
  this: Actor ⇒

  private[this] val actorClassName = this.getClass.getSimpleName

  private[this] var receiveName: String = {
    val receiveClassName = s"${this.receive.getClass.getSimpleName}"
    val left = receiveClassName.substring(0, receiveClassName.lastIndexOf("$"))
    left.substring(left.lastIndexOf("$") + 1)
  }

  def become(behavior: Actor.Receive): Unit = {
    val behaviorClassName = behavior.getClass.getSimpleName
    val left = behaviorClassName.substring(0, behaviorClassName.lastIndexOf("$"))

    receiveName = left.substring(left.lastIndexOf("$") + 1)

    context.become(behavior)
  }

  def info(message: Any): Unit = log.info(s"$actorClassName : $receiveName got $message")
}

然后，我的演员代码变成

class Hello extends Actor with MyLogging {

  def receive: Receive = behaviorFoo

  self ! Ping

  def behaviorFoo: Receive = {
    case any =>
      info(any)
      become(behaviorBar)
      self ! Ping
  }

  def behaviorBar: Receive = {
    case any => info(any)
  }
}

现在日志看起来像

...你好：behaviorFoo 得到了 Ping

...您好：behaviorBar 得到了 Ping

Answer 1

也许这就是那些被称为匿名函数的原因。 behaviorFoo、behaviorBar、... 名称对那些匿名函数没有影响。为了说明这一点，考虑这个：

// this is a val!
val behaviorFoo: Receive = {
  case _ =>
    log.info(this.receive.getClass.getSimpleName)
    become(behaviorBar)
    self ! Ping
}
// this returns the same anonymous function
def behaviorFoo2 = behaviorFoo

通过上面的内容，你应该看到你存储匿名函数的名称与匿名函数本身无关......

现在，如果你知道那些匿名函数是什么（它们是部分函数，别名类型为 Actor.Receive），你可以执行如下操作：

// not an anonymous function anymore
class BehaviourFoo extends Actor.Receive {
  val anonymousFun: Actor.Receive = {
    case _ =>
      log.info(this.receive.getClass.getSimpleName)
      become(behaviorBar)
      self ! Ping
  }
  override def isDefinedAt(x: Any) = anonymousFun.isDefinedAt(x)

  override def apply(x: Any) = anonymousFun(x)
}
// again, the name behaviorFoo doesn't matter
def behaviorFoo: Receive = new BehaviourFoo

这当然不值得麻烦，但它应该可以帮助您了解正在发生的事情。

作为当前 actor 接收函数名

akka actor current receive function name

akka

akka-actor