scala - 如何从广播的 Akka 流中获取订阅者和发布者？

Question

在使用更复杂的图表时，我无法将发布者和订阅者从我的流程中移除。我的目标是提供发布者和订阅者的 API，并在内部运行 Akka 流。这是我的第一次尝试，效果很好。

val subscriberSource = Source.subscriber[Boolean]
val someFunctionSink = Sink.foreach(Console.println)

val flow = subscriberSource.to(someFunctionSink)

//create Reactive Streams Subscriber
val subscriber: Subscriber[Boolean] = flow.run()

//prints true
Source.single(true).to(Sink(subscriber)).run()

但是对于更复杂的广播图，我不确定如何获取订阅者和发布者对象？我需要部分图吗？

val subscriberSource = Source.subscriber[Boolean]
val someFunctionSink = Sink.foreach(Console.println)
val publisherSink = Sink.publisher[Boolean]

FlowGraph.closed() { implicit builder =>
  import FlowGraph.Implicits._

  val broadcast = builder.add(Broadcast[Boolean](2))

  subscriberSource ~> broadcast.in
  broadcast.out(0) ~> someFunctionSink
  broadcast.out(1) ~> publisherSink
}.run()

val subscriber: Subscriber[Boolean] = ???
val publisher: Publisher[Boolean] = ???

Answer 1

当您调用 RunnableGraph.run() 时，流会运行，结果是该运行的“物化值”。

在您的简单示例中，Source.subscriber[Boolean] 的具体化值为 Subscriber[Boolean]。在您的复杂示例中，您希望将图形的多个组件的具体化值组合成一个元组 (Subscriber[Boolean], Publisher[Boolean]) 的具体化值。

您可以通过将您对其物化值感兴趣的组件传递给 FlowGraph.closed() 来实现，然后指定一个函数来组合物化值:

import akka.stream.scaladsl._
import org.reactivestreams._

val subscriberSource = Source.subscriber[Boolean]
val someFunctionSink = Sink.foreach(Console.println)
val publisherSink = Sink.publisher[Boolean]

val graph =
  FlowGraph.closed(subscriberSource, publisherSink)(Keep.both) { implicit builder ⇒
    (in, out) ⇒
      import FlowGraph.Implicits._

      val broadcast = builder.add(Broadcast[Boolean](2))

      in ~> broadcast.in
      broadcast.out(0) ~> someFunctionSink
      broadcast.out(1) ~> out
  }
val (subscriber: Subscriber[Boolean], publisher: Publisher[Boolean]) = graph.run()

见 Scaladocs for more information about the overloads of FlowGraph.closed .

(Keep.both 是函数的缩写 (a, b) => (a, b))