F# 在携带状态的同时绑定到输出

Question

我正在尝试使用计算表达式来构建操作列表。我需要绑定到从 getFood 操作返回的值，以便我可以注册稍后的步骤来使用它。

type Food =
    | Chicken
    | Rice

type Step =
    | GetFood of Food
    | Eat of Food
    | Sleep of duration:int

type Plan = Plan of Step list

type PlanBuilder () =

    member this.Bind (plan:Plan, f) =
        f plan
    member this.Yield _ = Plan []
    member this.Run (Plan x) = Plan (List.rev x)

    [<CustomOperation("eat")>]
    member this.Eat (Plan p, food) =
        printfn "Eat"
        Plan ((Eat food)::p)

    [<CustomOperation("sleep")>]
    member this.Sleep (Plan p, duration) =
        printfn "Sleep"
        Plan ((Sleep duration)::p)

let plan = PlanBuilder()

let rng = System.Random(123)


let getFood (Plan p) =
    printfn "GetFood"
    let randomFood = 
        if rng.NextDouble() > 0.5 then
            Food.Chicken
        else
            Food.Rice
    (Plan ((GetFood randomFood)::p)), randomFood

let testPlan =
    plan {
        let! food = getFood // <-- This is what I am trying to get to work
        sleep 10
        eat food
    }

我认为问题出在 Bind 但我不知道是什么。

(*
Example result
testPlan =
    (GetFood Chicken,(
        (Sleep 10,(
            EatFood Chicken
        ))
    ))
*)

Answer 1

要完成这样的工作，您可能需要一种具有更多单子结构并允许您存储任何结果的类型，而不仅仅是计划。我会使用这样的东西：

type Step =
  | GetFood of Food
  | Eat of Food
  | Sleep of duration:int

type Plan<'T> = Plan of Step list * 'T

现在，Plan<'T> 表示生成类型 'T 的值并沿途收集计划的计算。 GetFood可以出方案，还return菜：

let getFood () =
  printfn "GetFood"
  let randomFood = 
    if rng.NextDouble() > 0.5 then Food.Chicken
    else Food.Rice
  Plan([GetFood randomFood], randomFood)

实施计算构建器有点像魔法，但您现在可以定义 Bind 和您的自定义操作。为了能够访问参数中的变量，它需要是一个函数，如 where 或 select 操作：

type PlanBuilder () =

  member this.For (Plan(steps1, res):Plan<'T>, f:'T -> Plan<'R>) : Plan<'R> =
    let (Plan(steps2, res2)) = f res
    Plan(steps1 @ steps2, res2)

  member this.Bind (Plan(steps1, res):Plan<'T>, f:'T -> Plan<'R>) : Plan<'R> =
      let (Plan(steps2, res2)) = f res
      Plan(steps1 @ steps2, res2)
  
  member this.Yield x = Plan([], x)
  member this.Return x = Plan([], x)
  member this.Run (Plan(p,r)) = Plan(List.rev p, r)

  [<CustomOperation("eat", MaintainsVariableSpace=true)>]
  member this.Eat (Plan(p, r), [<ProjectionParameter>] food) =
      Plan((Eat (food r))::p, r)

  [<CustomOperation("sleep", MaintainsVariableSpace=true)>]
  member this.Sleep (Plan(p, r), [<ProjectionParameter>] duration) =
      Plan ((Sleep (duration r))::p, r)

let plan = PlanBuilder()

这实际上可以让您实施测试计划：

let testPlan =
  plan {
      let! food = getFood () 
      sleep 10
      eat food
      return ()
  }

也就是说，在实践中，我不确定我是否真的想要使用它。我可能只会使用 seq { .. } 计算，使用 yield 来累积计划的步骤。

F# 在携带状态的同时绑定到输出

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