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我对 Akka 世界很陌生，更不用说 Akka 类型了。如果这个人很愚蠢，也很抱歉，:(

所以，我遵循了请求-响应交互模式，让参与者进行交互。这意味着在我的命令中添加一个 actorRef。

但是我的命令都在 protobuf 中。我不得不序列化actorRef。在不知道actor2的ActorSystem的情况下，actor1如何反序列化actor2的actorRef？

我读过一个答案：（https://manuel.bernhardt.io/2018/07/20/akka-anti-patterns-java-serialization/#comment-157564）但我无法理解解决方案......： (

WorkgroupCommand.proto

message WorkgroupCommand {
    oneof sealed_value {
        EnqueueWorkDone enq = 1;
        QueueFull qF = 2;
    }
}

message EnqueueWorkDone {
    required string id = 1;
    required string replyTo = 2; //serialized actorRef
}

工作组.scala

private val system1 = ActorSystem(Behaviors.empty, "system-1")

val commandHandler:(State, WorkgroupCommand) => Effect[Event,State] ={
  case (state, EnqueueWorkDone(id, replyTo)) =>
     //how to deserialize replyTo ?????

}

代理.scala

private val system2 = ActorSystem(mainBehavior, "system-2")
private val resolver = ActorRefResolver(system2.toTyped)

def mainBehavior = Behaviors.setup{ context => 
  //assuming the we have the Workgroup actor
  Workgroup ! EnqueueWorkDone("12345",
                resolver.toSerializationFormat(context.self))

  Behaviors.same
}

有办法吗？如果 Workgroup 和 Agent 都是主要 actorSystem 的子代，会有帮助吗？

4

1 回答 1

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val actorRefResolver = ActorRefResolver(system.toTyped)

and then

val serializedActorRef: Array[Byte]= actorRefResolver.toSerializationFormat(replyTo).getBytes(StandardCharsets.UTF_8)

val str = new String(serializedActorRef, StandardCharsets.UTF_8)

val deserializedActorRef = actorRefResolver.resolveActorRef[SomeType.type](str)
于 2021-07-20T11:08:49.110 回答