Question

我有这样的东西：

implicit val system = ActorSystem("actorSystem")
implicit val materializer = ActorMaterializer()

val (ref: ActorRef, publisher: Publisher[(String, String, String)]) =
  Source.actorRef[(String, String, String)](bufferSize = 1000, OverflowStrategy.fail)
    .toMat(transferFTP(_))(Keep.both)
    .run()

def transferFTP(data: (String, String, String)): Sink[ByteString, Future[IOResult]] = {
  // ...
}

在我阅读的每个示例中，toMat()都使用Sink作为参数，但是编译器期望Graph[SinkShape[(String,String,String)], NotInferedMat2]。

Answer 1

toMat确实需要Sink。 Sink[(String, String, String), _]是Graph[SinkShape[(String, String, String)], _]（换句话说，前者是后者的子类）。问题之一是您传递的是Sink[ByteString, _]。此外，当您的Publisher[(String, String, String)]的物化值为Sink时，您期望的是Future[IOResult]。

下面是一个类型对齐的示例，从而使编译器感到满意：

val tupleToByteString: ((String, String, String)) => ByteString = ???

val transferFTP: Sink[ByteString, Future[IOResult]] = ???

val (ref: ActorRef, sink: Future[IOResult]) =
  Source.actorRef[(String, String, String)](bufferSize = 1000, OverflowStrategy.fail)
    .via(Flow.fromFunction(tupleToByteString)) // alternatively: .map(tupleToByteString)
    .toMat(transferFTP)(Keep.both)
    .run()

在Akka Streams中将Sink转换为Graph（SinkShape）？

1 个答案: