Question

我打算创建一个自定义路由器，将消息传递给未知数量的动态创建的actor（所有这些）。因此，我需要维护一个类似于here所示想法的“订阅者”列表。但我似乎无法通过上下文切换维护列表。我无法理解为什么创建的可变哈希集在上下文切换后被重置。有办法规避吗？代码：

class myRouter extends Actor {
 var subscribers: mutable.HashSet[ActorRef]()
 var waitingFor = subscribers

 def receive = receiveSubAndRequests

 def receiveSubAndRequests: Actor.Receive = {

  case RegisterWorker =>
   subscribers += sender() //this happens right after initiation
  case e: ComputationParams => //say these are guaranteed to come after all have subscribed
   subscribers.foreach(s=> s.tell(ComputeIt))//perform some compute. this works!
   context.become(receiveResults(e.foo, sender())//pass some state
 }  

 def receiveResults(num: Int, sendTo: ActorRef): Actor.Receive = {

  case result: Result =>
   waitingFor -= sender() //this works, meaning var was accessed
   //other stuff until all results are in

  case foo: OtherParams => //say we need to handle and initiate another ComputeIt
    // stuff
    subscribers.foreach(s => s.tell(ComputeIt)//!!subscribers is empty!!
    //other
  }

Answer 1

您的waitingFor字段只是设置为指向subscribers的指针，因此当执行第waitingFor -= sender()行时，发件人将被有效地从subscribers中移除（正在与指向waitingFor）相同的可变HashSet。我认为你要找的是在开始计算时初始化waitingFor：

case e: ComputationParams => //say these are guaranteed to come after all have subscribed
  subscribers.foreach(s=> s.tell(ComputeIt))//perform some compute. this works!
  waitingFor = new mutable.HashSet(subscribers) // ***
  context.become(receiveResults(e.foo, sender())//pass some state

在标记为***的行上，您（重新）创建waitingFor为新 HashSet，初始化为保留所有相同的发件人值subscribers HashSet。当你在＆＃34; receiveResults＆＃34;中处理case foo: OtherParams =>时，你也可能需要这样做。状态。

Akka - 如何通过上下文切换来维护状态

1 个答案: