Question

我创建了以下测试代码-.Net core 2.1控制台应用程序。它仅打印以下消息

TestActor从[akka：// MySystem / user / Scheduler＃1426101451]收到消息MyTask（“ Test1”，“ Test1”）

但是演员Ok 0无法收到消息scheduler吗？

open System
open Akka.FSharp
open Akka

type MyTask = MyTask of item1: string * item2: string 

let system = System.create "MySystem" <| Configuration.load ()

let scheduler (actors: Actor.IActorRef) (mailbox: Actor<Result<int, string>>) =
    let rec loop (list: int list list) = actor {
        let! m = mailbox.Receive ()
        let sender = mailbox.Sender ()
        let akkaName = mailbox.Self.Path.Name
        printfn "%s received message %A from %A" akkaName m sender
        return! loop []
    }
    actors <! MyTask("Test1", "Test1")
    loop []

let processor (mailbox: Actor<MyTask>) =
    let rec loop () = actor {
        let! m = mailbox.Receive ()
        let sender = mailbox.Sender ()
        let akkaName = mailbox.Self.Path.Name
        printfn "%s received message %A from %A" akkaName m sender
        sender <! Ok 0 // scheduler cannot receive this message?
        return! loop ()
    }
    loop ()

[<EntryPoint>]
let main argv =
    let actor = spawn system "TestActor" processor
    spawn system "Scheduler" (scheduler actor) |> ignore
    system.WhenTerminated.Wait()
    0

更新：

在我将参数从(mailbox: Actor<Result<int, string>>)更改为(mailbox: Actor<_>)后可以使用吗？

Answer 1

scheduler不是MyTask所回复的processor消息的发件人，因为您是从参与者外部进行告诉（<!）计算。这意味着它基本上是在没有发件人的情况下发送的。您可以在Tell上使用IActorRef方法与显式发送者一起发送，因为您想从actor计算的上下文之外发送它：

actors.Tell(MyTask("Test1", "Test1"), mailbox.Self)

编辑

另一个问题是mailbox的{{1}}参数键入为scheduler，但是由于在您的上下文中Actor<Result<int, string>>会被推断为Ok 0，它与actor类型签名不匹配，该消息将被忽略。当使用特定消息类型定义参与者时，任何其他类型的消息都将被忽略。

Actor无法接收发送到`mailbox.Sender（）`的消息吗？

更新：

1 个答案: