调用组合时的 F# 计算表达式

Question

我试图直接实现 Maybe单子(monad)。所以基本上整个表达式的计算结果为 Nothing如果中间步骤之一是 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
    }

我期待 foobar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
    }

但是随后您需要将其定义添加到构建器中:

member this.ReturnFrom (expr) = expr

然后编译器会要求您添加一个延迟方法，在您的情况下，我认为您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 Combine您定义的两个参数没有使用相同的类型，您可以修复它或在返回类型中添加类型注释:

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

打印和:

val foobar : Maybe<int> = Nothing

如果您想了解 CE 的所有详细信息，请查看这篇不错的文章: https://www.microsoft.com/en-us/research/publication/the-f-computation-expression-zoo/