你如何创建一个F#工作流来支持单步操作？

我想创建一个构建器，它构建的表达式在每个步骤后都会返回类似于continuation的内容

大概是这样的：

module TwoSteps = 
  let x = stepwise {
    let! y = "foo"
    printfn "got: %A" y
    let! z = y + "bar"
    printfn "got: %A" z
    return z
  }

  printfn "two steps"
  let a = x()
  printfn "something inbetween"
  let b = a()

其中“let a”行返回包含后面要计算的其余表达式的内容

对于每个步骤数，使用单独的类型执行此操作非常简单，但当然不是特别有用：

type Stepwise() =
  let bnd (v: 'a) rest = fun () -> rest v
  let rtn v = fun () -> Some v
  member x.Bind(v, rest) = 
    bnd v rest
  member x.Return v = rtn v

let stepwise = Stepwise()

module TwoSteps = 
  let x = stepwise {
    let! y = "foo"
    printfn "got: %A" y
    let! z = y + "bar"
    printfn "got: %A" z
    return z
  }

  printfn "two steps"
  let a = x()
  printfn "something inbetween"
  let b = a()

module ThreeSteps = 
  let x = stepwise {
    let! y = "foo"
    printfn "got: %A" y
    let! z = y + "bar"
    printfn "got: %A" z
    let! z' = z + "third"
    printfn "got: %A" z'
    return z
  }

  printfn "three steps"
  let a = x()
  printfn "something inbetween"
  let b = a()
  printfn "something inbetween"
  let c = b()

结果就是我想要的：

two steps
got: "foo"
something inbetween
got: "foobar"
three steps
got: "foo"
something inbetween
got: "foobar"
something inbetween
got: "foobarthird"

但我不知道一般情况会是什么

我希望能够将事件反馈到此工作流中，这样您就可以编写如下内容：

let someHandler = Stepwise<someMergedEventStream>() {
  let! touchLocation = swallowEverythingUntilYouGetATouch()
  startSomeSound()
  let! nextTouchLocation = swallowEverythingUntilYouGetATouch()
  stopSomeSound()
}

让someHandler=Stepwise（）{
let！touchLocation=吞下所有东西直到UgetaTouch（）
startSomeSound（）
让！nextTouchLocation=吞下所有东西直到UgetTouch（）
停止声音
}

---

并让事件触发工作流中的下一步。（特别是，我想在iPhone上用MonoTouch-F#玩这种东西。传递objc选择器让我发疯。）

Monads和计算构建器把我搞糊涂了，但我已经改编了我在一个。也许有些零碎的东西会有用

下面的代码包含一个操作队列和一个表单，单击事件在其中侦听操作队列中可用的下一个操作。下面的代码是一个连续执行4个操作的示例。在FSI中执行它并开始单击表单

open System.Collections.Generic
open System.Windows.Forms

type ActionQueue(actions: (System.EventArgs -> unit) list) =
    let actions = new Queue<System.EventArgs -> unit>(actions) //'a contains event properties
    with
        member hq.Add(action: System.EventArgs -> unit) = 
           actions.Enqueue(action)
        member hq.NextAction = 
            if actions.Count=0 
                then fun _ -> ()
                else actions.Dequeue()

//test code
let frm = new System.Windows.Forms.Form()

let myActions = [
    fun (e:System.EventArgs) -> printfn "You clicked with %A" (e :?> MouseEventArgs).Button
    fun _ -> printfn "Stop clicking me!!"
    fun _ -> printfn "I mean it!"
    fun _ -> printfn "I'll stop talking to you now."
    ]

let aq = new ActionQueue(myActions)

frm.Click.Add(fun e -> aq.NextAction e)

//frm.Click now executes the 4 actions in myActions in order and then does nothing on further clicks
frm.Show()

---

您的实现的问题在于，它为每个要绑定的调用返回“unit->'a”，因此对于不同的步骤数，您将得到不同类型的结果（通常，这是monad/计算表达式的可疑定义）

---

正确的解决方案应该是使用其他类型，它可以表示具有任意步数的计算。您还需要区分两种类型的步骤——有些步骤只计算计算的下一步，有些步骤返回结果（通过

return

关键字）。我将使用类型

seq在某种程度上，我试图得到的是使用工作流来构建手工操作的东西。工作流将生成一个函数序列，然后您只需迭代该序列。只是一个补充，这也非常类似于所谓的恢复单子。但是我相信在F#中使用序列的实现可能会更容易一些。“PS：我发现这个问题非常有趣！你介意我将我的答案添加到我的博客（）的博客帖子中吗？”当然，请随意。谢谢你的回答；我现在正在研究如何将其应用到iPhone事件系统中。
let moreActions = [
    fun _ -> printfn "Ok, I'll talk to you again. Just don't click anymore, ever!"
    fun _ -> printfn "That's it. I'm done with you."
    ]

moreActions |> List.iter (aq.Add)

let foo() = stepwise { 
  return 1; }
let bar() = stepwise { 
  let! a = foo()
  return a + 10 }

type Stepwise() = 
  member x.Bind(v:seq<option<_>>, rest:(_ -> seq<option<_>>)) = seq {
    let en = v.GetEnumerator()
    let nextVal() = 
      if en.MoveNext() then en.Current
      else failwith "Unexpected end!" 
    let last = ref (nextVal())
    while Option.isNone !last do
      // yield None for each step of the source 'stepwise' computation
      yield None
      last := next()
    // yield one more None for this step
    yield None      
    // run the rest of the computation
    yield! rest (Option.get !last) }
  member x.Return v = seq { 
    // single-step computation that yields the result
    yield Some(v) }

let stepwise = Stepwise() 
// simple function for creating single-step computations
let one v = stepwise.Return(v)

let oneStep = stepwise {
  // NOTE: we need to explicitly create single-step 
  // computations when we call the let! binder
  let! y = one( "foo" ) 
  printfn "got: %A" y 
  return y + "bar" } 

let threeSteps = stepwise { 
  let! x = oneStep // compose computations :-)
  printfn "got: %A" x 
  let! y = one( x + "third" )
  printfn "got: %A" y
  return "returning " + y } 

for step, idx in Seq.zip threeSteps [ 1 .. 10] do
  printf "STEP %d: " idx
  match step with
  | None _ -> ()
  | Some(v) -> printfn "Final result: %s" v