Question

我试图实现这个直接Maybe monad。因此，如果其中一个中间步骤是Nothing，那么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
    }

我希望将foobar翻译为Just "foo" >>= fun foo -> Combine(Just 1, Nothing)，但不会调用Combine。

Answer 1

这不是计算表达式的编写方式。每次你想要'产生结果'时你需要在表达式的左侧添加一些关键字（return，return！，yield或yield！），在你的例子中我会添加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Delay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()

几乎在那里，现在你有一个值限制，很可能是因为你定义的Combine在两个参数上都没有使用相同的类型，你可以修复它或只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，现在你得到了：

Bind called
Combine called

打印并且：

val foobar : Maybe<int> = Nothing

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

调用Combine时的F＃计算表达式

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