调用组合时的 F# 计算表达式

Question

我试图直接实现这个Maybe monad。因此，如果中间步骤之一是Nothing，则基本上整个表达式的计算结果为Nothing。

type Maybe<'a> =
    | Just of 'a
    | Nothing
type MaybeBuilder () =
    member this.Combine ((first, second) : Maybe<'a> * Maybe<'b>) : Maybe<'b> =
        printfn "Combine called"
        match first with
        | Nothing -> Nothing
        | _ ->
            match second with
            | Nothing -> Nothing
            | _ as a -> a
     member this.Zero () = Just ()
     member this.Bind((m, f) : Maybe<'a> * ('a -> Maybe<'b>)) =
        printfn "Bind called"
        match m with
        | Nothing -> Nothing
        | Just a -> f a
let MaybeMonad = MaybeBuilder()
let foobar =
    MaybeMonad {
        let! foo = Just "foo"
        Just 1
        Nothing
    }

我预计foobar 会被翻译成Just "foo" >>= fun foo -> Combine(Just 1, Nothing)，但是Combine 没有被调用。

Answer 1

这不是计算表达式的预期编写方式。每次您想要“产生结果”时，您都需要在表达式的左侧添加一些关键字（return、return！、yield 或 yield！），在您的示例中，我将添加一个 return!：

let foobar =
    MaybeMonad {
        let! foo = Just "foo"
        return! Just 1
        return! Nothing
    }

但是你需要将它的定义添加到构建器中：

member this.ReturnFrom (expr) = expr

然后编译器会要求你添加一个延迟方法，在你的情况下，我认为你正在寻找类似的东西：

member this.Delay(x) = x()

差不多了，现在你有一个值限制，很可能是因为你定义的Combine 没有在两个参数上使用相同的类型，你可以修复它或者只是在返回类型中添加一个类型注释：

let foobar : Maybe<int> =
    MaybeMonad {
        let! foo = Just "foo"
        return! Just 1
        return! Nothing
    }

就是这样，现在你得到：

Bind called
Combine called

印刷和：

val foobar : Maybe<int> = Nothing

如果您想了解 CE 的所有细节，请查看这篇精彩的文章：https://www.microsoft.com/en-us/research/publication/the-f-computation-expression-zoo/