f# - 如何避免对 Actor 实例的可变引用

Question

我正尝试在 F# 中获得一些使用 Akka.NET actors 的经验。我有一个场景，当一个 Actor 需要将另一个 Actor 作为他的 child 时。第一个 Actor 将转换每条消息，然后将结果发送给另一个 Actor 。我使用 actorOf2 函数生成 Actor 。这是我的代码:

let actor1 work1 work2 =
  let mutable actor2Ref = null
  let imp (mailbox : Actor<'a>) msg =
    let result = work1 msg
    if actor2Ref = null then 
      actor2Ref <- spawn mailbox.Context "decide-actor" (actorOf2 <| work2)
    actor2Ref <! result
  imp

let actor1Ref = actor1 work1' work2'
  |> actorOf2 
  |> spawn system "my-actor"

我不喜欢的是可变的 Actor 引用。但是我必须让它可变，因为我需要 mailbox.Context 来生成一个子 actor，而且在第一次调用之前我没有任何上下文。我看到了this question但我不知道如何将它应用到我的函数中。

在更高级的场景中，我需要一组按键分区的子 actor。在这种情况下，我使用的是 Actor 引用词典。有更好的(更像 F# 的)方法吗？

Answer 1

为了在迭代中保持您的“状态”，您需要使迭代明确。这样，您可以将当前“状态”作为尾调用参数传递。就像您链接的问题一样:

let actor1 work1 work2 (mailbox : Actor<'a>) =
  let rec imp actor2 =
    actor {
      let! msg = mailbox.Receive()
      let result = work1 msg

      let actor2 =
        match actor2 with
        | Some a -> a // Already spawned on a previous iteration
        | None -> spawn mailbox.Context "decide-actor" (actorOf2 <| work2)

      actor2 <! result
      return! imp (Some actor2)
    }

  imp None

现在，您不需要使用 actorOf2 或 actorOf 来生成这个 actor，因为它已经有了正确的签名:

let actor1Ref = 
  actor1 work1' work2'
  |> spawn system "my-actor"

.
编辑
如果您担心额外的样板文件，没有什么能阻止您将样板文件作为函数打包(毕竟，actorOf2 does something similar ):

let actorOfWithState (f: Actor<'msg> -> 'state -> 'msg -> 'state) (initialState: 'state) mailbox =
  let rec imp state =
    actor {
      let! msg = mailbox.Receive()
      let newState = f mailbox state msg
      return! imp newState
    }

  imp initialState

然后:

let actor1 work1 work2 (mailbox : Actor<'a>) actor2 msg =
  let result = work1 msg
  let actor2 =
    match actor2 with 
    | Some a -> a
    | None -> spawn mailbox.Context "decide-actor" (actorOf2 work2)

  actor2 <! result
  actor2

let actor1Ref = 
  actor1 work1' work2'
  |> actorOfWithState
  |> spawn system "my-actor"