参与者无法接收发送到`mailbox.Sender ()`的消息？

Question

我创建了以下测试代码- .Net核心2.1控制台应用程序。它只打印以下消息

TestActor从akka接收到消息MyTask ("Test1"，"Test1")：//MySystem/user/Scheduler#1426101451

但是，参与者Ok 0不能接收消息scheduler。

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：

当我将参数从(mailbox: Actor<Result<int, string>>)更改为(mailbox: Actor<_>)后，它就可以工作了。

Answer 1

scheduler不是processor响应的MyTask消息的发送方，因为您正在从参与者计算之外执行tell (<!)操作。这意味着它基本上是在没有发送者的情况下发送的。您可以使用Tell方法在IActorRef上与显式发送方一起发送，因为您希望从参与者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参数的scheduler被键入为Actor<Result<int, string>>，但是由于在上下文中Ok 0将被推断为Result<int,obj>，因此它将与参与者类型签名不匹配，并且消息将被忽略。当使用特定消息类型定义参与者时，任何不同类型的消息都将被忽略。