实现指树时键入错误

pri*_*tor 7 recursion f# types infinite finger-tree

我想用2-3个手指树来玩,如Hinze的(Haskell)论文所述(另见本博客).

type Node<'a> =
    | Node2 of 'a * 'a
    | Node3 of 'a * 'a * 'a

    static member OfList = function
        | [a; b] -> Node2(a, b)
        | [a; b; c] -> Node3(a, b, c)
        | _ -> failwith "Only lists of length 2 or 3 accepted!"

    member me.ToList () =
        match me with
        | Node2(a, b) -> [a; b]
        | Node3(a, b, c) -> [a; b; c]

type Digit<'a> =
    | One of 'a
    | Two of 'a * 'a
    | Three of 'a * 'a * 'a
    | Four of 'a * 'a * 'a * 'a

    static member OfList = function
        | [a] -> One(a)
        | [a; b] -> Two(a, b)
        | [a; b; c] -> Three(a, b, c)
        | [a; b; c; d] -> Four(a, b, c, d)
        | _ -> failwith "Only lists of length 1 to 4 accepted!"

    member me.ToList () =
        match me with
        | One a -> [a]
        | Two(a, b) -> [a; b]
        | Three(a, b, c) -> [a; b; c]
        | Four(a, b, c, d) -> [a; b; c; d]

    member me.Append x =
        match me with
        | One a -> Two(a, x)
        | Two(a, b) -> Three(a, b, x)
        | Three(a, b, c) -> Four(a, b, c, x)
        | _ -> failwith "Cannot prepend to Digit.Four!"

    member me.Prepend x =
        match me with
        | One a -> Two(x, a)
        | Two(a, b) -> Three(x, a, b)
        | Three(a, b, c) -> Four(x, a, b, c)
        | _ -> failwith "Cannot prepend to Digit.Four!"

[<NoComparison>]
[<NoEquality>]
type FingerTree<'a> =
    | Empty
    | Single of 'a
    | Deep of Digit<'a> * FingerTree<Node<'a>> * Digit<'a>

type Digit<'a> with
    member me.Promote () =
        match me with
        | One a -> Single a
        | Two(a, b) -> Deep(One a, Empty, One b)
        | Three(a, b, c) -> Deep(One a, Empty, Two(b, c))
        | Four(a, b, c, d) -> Deep(Two(a, b), Empty, Two(c, d))

type View<'a> = Nil | View of 'a * FingerTree<'a>
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现在我只是无法使viewl功能工作,它抱怨类型不匹配:

期待一个FingerTree <'a>但是给了一个FingerTree>.

统一''a'和'Node <'a>'FingerTree时,结果类型将是无限的.

let rec viewl : FingerTree<'a> -> View<'a> = function
    | Empty -> Nil
    | Single x -> View(x, Empty)
    | Deep(One x, deeper(*:FingerTree<'a>/FingerTree<Node<'a>>*), suffix) ->
        let rest =
            match viewl deeper with
            | Nil ->
                suffix.Promote()
            | View (node(*:Node<'a>*), rest) ->
                let prefix = node.ToList() |> Digit<_>.OfList
                Deep(prefix, rest, suffix)
        View(x, rest)
    | Deep(prefix, deeper, suffix) ->
        match prefix.ToList() with
        | x::xs ->
            View(x, Deep(Digit<_>.OfList xs, deeper, suffix))
        | _ -> failwith "Impossible!"
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我之前有这个错误,prepend但能够通过添加函数的完整类型信息来解决它.

// These three/four type annotations solved the problem.
let rec prepend<'a> (a:'a) : FingerTree<'a> -> FingerTree<'a> = function
    | Empty -> Single a
    | Single b -> Deep(One a, Empty, One b)
    | Deep(Four(b, c, d, e), deeper, suffix) ->
        Deep(Two(a, b), prepend (Node3(c, d, e)) deeper, suffix)
    | Deep(prefix, deeper, suffix) ->
        Deep(prefix.Prepend a, deeper, suffix)
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因为viewl这似乎不够,所以我也尝试在函数中间添加类型(查找注释).它没用.

我有点理解错误及其来源.谁能告诉我如何使这个工作?恕我直言,应该是可能的,因为否则prepend也不会编译.也许像这样的技巧有帮助吗?(虽然不明白).


PS:我还把代码放在FsSnip上,以便在浏览器中播放.

kvb*_*kvb 5

函数喜欢viewlprepend依赖于多态递归:递归调用prepend采用与原始调用不同类型的参数.您可以在F#中定义此类函数,但正如您发现它们需要完整的类型注释(否则您会收到非常混乱的错误消息).特别要注意,类型参数必须在函数的定义中是显式的(尽管通常可以在调用站点推断它们).所以第一个问题是你需要viewl<'a>在定义中指定.

然而,这是一个非常微妙的第二个问题,这是一个问题Digit<_>.OfList.尝试将第一块代码发送到F#interactive并查看结果定义的签名:您将看到static member OfList : (obj list -> Digit<obj>),随后将使其viewl无法正确定义.所以发生了什么事?你没有签名OfList,所以它不是一般的方法(函数将被推广,但成员永远不会).但是编译器还不能推断出你打算输入列表为类型'a list,其中'a是该类型的泛型参数-为什么会推断出这个特定的类型,而不是int liststring list等?所以它选择了一个无聊的单态默认值(obj list),除非你在后续代码中做一些事情来将它约束到一个不同的具体单态类型.相反,你需要添加一个签名Digit,然后一切都会好的.

通常在F#中,为每个类型创建一个单独的模块以定义相关函数等是惯用的ToList.由于函数定义是一般化的,这也可以避免Digit您在这里遇到的问题.也就是说,您可以像这样构建代码:

type Node<'a> =
    | Node2 of 'a * 'a
    | Node3 of 'a * 'a * 'a

module Node =
    let ofList = function
    | [a; b] -> Node2(a, b)
    | [a; b; c] -> Node3(a, b, c)
    | _ -> failwith "Only lists of length 2 or 3 accepted!"

    let toList = function
    | Node2(a, b) -> [a; b]
    | Node3(a, b, c) -> [a; b; c]

type Digit<'a> =
    | One of 'a
    | Two of 'a * 'a
    | Three of 'a * 'a * 'a
    | Four of 'a * 'a * 'a * 'a

[<NoComparison>]
[<NoEquality>]
type FingerTree<'a> =
    | Empty
    | Single of 'a
    | Deep of Digit<'a> * FingerTree<Node<'a>> * Digit<'a>

module Digit =
    let ofList = function
    | [a] -> One(a)
    | [a; b] -> Two(a, b)
    | [a; b; c] -> Three(a, b, c)
    | [a; b; c; d] -> Four(a, b, c, d)
    | _ -> failwith "Only lists of length 1 to 4 accepted!"

    let toList = function
    | One a -> [a]
    | Two(a, b) -> [a; b]
    | Three(a, b, c) -> [a; b; c]
    | Four(a, b, c, d) -> [a; b; c; d]

    let append x = function
    | One a -> Two(a, x)
    | Two(a, b) -> Three(a, b, x)
    | Three(a, b, c) -> Four(a, b, c, x)
    | _ -> failwith "Cannot prepend to Digit.Four!"

    let prepend x = function
    | One a -> Two(x, a)
    | Two(a, b) -> Three(x, a, b)
    | Three(a, b, c) -> Four(x, a, b, c)
    | _ -> failwith "Cannot prepend to Digit.Four!"

    let promote = function
    | One a -> Single a
    | Two(a, b) -> Deep(One a, Empty, One b)
    | Three(a, b, c) -> Deep(One a, Empty, Two(b, c))
    | Four(a, b, c, d) -> Deep(Two(a, b), Empty, Two(c, d))

type View<'a> = Nil | View of 'a * FingerTree<'a>

let rec viewl<'a> : FingerTree<'a> -> View<'a> = function
    | Empty -> Nil
    | Single x -> View(x, Empty)
    | Deep(One x, deeper, suffix) ->
        let rest =
            match viewl deeper with
            | Nil -> suffix |> Digit.promote
            | View (node, rest) ->
                let prefix = node |> Node.toList |> Digit.ofList
                Deep(prefix, rest, suffix)
        View(x, rest)
    | Deep(prefix, deeper, suffix) ->
        match prefix |> Digit.toList with
        | x::xs ->
            View(x, Deep(Digit.ofList xs, deeper, suffix))
        | _ -> failwith "Impossible!"
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  • 不,函数和成员都可以使用多态递归,您只需要确保完全注释定义.你的代码几乎完全没问题 - 只需进行我在前两段中提到的小改动.我在底部插入代码以显示更惯用的方法,但如果您对现有结构感到满意,则无需这样做. (2认同)