我有点新akka打字，我正试图发送一个消息，需要在给定时间内回答.
我发现了带有ask的请求-响应模式，这看起来很有趣，但是有没有一种方法可以在已经定义的Behaviours.receive中实现它？
这里的想法是在每次玩家应答时或超时后调用nextPlayerTurn

override def refereeTurn(): Behavior[Msg] = Behaviors.receive {
   case (_, msg: GuessMsg) =>
      if(currentPlayer.isDefined && currentPlayer.get == msg.getSender) {
        controller ! msg
      } else {
        println("Player tried to guess after Timeout")
      }
      Behaviors.same
  case (context, msg: ReceivedResponseMsg) =>
      if(currentPlayer.isDefined && currentPlayer.get == msg.getSender) 
        nextPlayerTurn(context)     
      Behaviors.same
      ...
}

...

/**
   * Tells to a player to start his turn and sets a timer that defines time in which a player has to make a guess.
   * If such guess isn't made, sends that user an end turn message, fails the promise of his turn and allows next
   * player to play his turn

* /

  override def nextPlayerTurn(ctx: ActorContext[Msg]): Unit = {
    implicit val timeout: Timeout = Timeout.timeout
    currentPlayer = Option(turnManager.nextPlayer)
    ctx.ask[Msg,Msg](currentPlayer.get, ref => YourTurnMsg(ref)) {
      case Success(msg: GuessMsg) => println("\n SUCCESS"); msg
      case Failure(_) => println(currentPlayer.get +" didn't guess in time"); TurnEnd(currentPlayer.get)
      case _ => TurnEnd(currentPlayer.get)
    }
  }

在这种情况下，在发送YourTurnMsg后，玩家应该使用停止计时器的GuessMsg进行响应，这从未发生，因为正在执行refereeTurn Begaviour内部的大小写匹配，而不是Success（而在超时后总是给出Failure）。
我是不是对询问模式有了错误的想法，应该用计时器来做一个新的行为？