F# 计算表达式：何时调用 Combine？

Question

F# 计算表达式：何时调用 Combine？

6

我尝试实现了这个简单的Maybe Monad，如果其中一个中间步骤为Nothing，则整个表达式将被评估为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
    }

我原本期望 foobar 被翻译为 Just "foo" >>= fun foo -> Combine(Just 1, Nothing)，但实际上并没有调用 Combine。

- sqd

1个回答

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- Gus · Accepted Answer

计算表达式的书写方式不应该是这样的。每次想要“产生结果”时，需要在表达式的左侧添加一些关键词（return、return!、yield或yield!）。在你的例子中，我会添加一个return!：

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

但是您需要将其定义添加到构建器中：

member this.ReturnFrom (expr) = expr

如果编译器要求您添加Delay方法，则在您的情况下，我认为您正在寻找类似于以下内容的内容：

member this.Delay(x) = x()

离目标已经很接近了，现在您遇到了一个值限制错误，很可能是因为您定义的Combine没有在两个参数上使用相同的类型，您可以修复它或只是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，现在你可以得到：

Bind called
Combine called

打印和：

val foobar : Maybe<int> = Nothing

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