Scala Akka Play， future 不会回来

Question

我正在使用 Scala 2.10、Akka 2.1 和 Play 2.1。当我向我的后端发送一个 http 请求时，我要求一个参与者计算一些东西。这个想法是如果它在超时之前返回，则返回计算结果，否则返回一个不同的字符串。请参阅下面的代码。

val futureInt: Future[Int] = ask(testActor, Calculate(number.toInt)).mapTo[Int]
val timeoutFuture = play.api.libs.concurrent.Promise.timeout("Oops", 2.seconds)

Async {
  Future.firstCompletedOf(Seq(futureInt, timeoutFuture)).map {
    case i: Int => Ok("Got result " + i)
    case t: String => Ok("timeout expired")
  }
}

Actor 如下:

class TestActor() extends Actor {
  def receive = {
    case Calculate(tonumber: Int) =>
      for (a <- 1 to tonumber) {
        val c: Double = scala.math.pow(a, 2)
        println("a: " + a + ", c: " + c)
      }
      12345 // hardcoded value to return when the calculation finishes
    case _ =>
      println("bah")
  }
}

我的问题是，即使 actor 在超时之前完成，Future 也不会“返回”任何内容，因此超时总是会过期。我究竟做错了什么？非常感谢。

Answer 1

来自 http://doc.akka.io/docs/akka/snapshot/scala/actors.html

Using ask will send a message to the receiving Actor as with tell, and the receiving actor must reply with sender ! reply in order to complete the returned Future with a value.

和

Warning

To complete the future with an exception you need send a Failure message to the sender. This is not done automatically when an actor throws an exception while processing a message.

因此，不要像在通常的 Scala 函数中那样“返回”，而是按照以下方式做一些事情

def receive = {
  case Calculate(tonumber: Int) =>
    ...
    sender ! 12345
  case _ =>
    sender ! akka.actor.Status.Failure(new InvalidArgumentException)
}