序列表达式和多态递归如何一起发挥作用？

Question

This project really is a questions 对我来说。

我已经了解了多态递归并且我明白为什么它是一种特殊情况，因此 F# 需要完整的类型注释。

对于常规函数，我可能需要一些操作，但通常都能正确完成。现在，我正在尝试将（工作中的）基本 toSeq 改编为更专业的手指树，但做不到。

我的感觉是计算表达式的使用有点关系。这是精简版：

module ThisWorks =

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

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

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

        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]

    module FingerTree =
        open Node
        open Digit

        type FingerTree<'a> =
            | Empty
            | Single of 'a
            | Deep of Digit<'a> * Lazy<FingerTree<Node<'a>>> * Digit<'a>

        let rec toSeq<'a> (tree:FingerTree<'a>) : seq<'a> = seq {
            match tree with
            | Single single ->
                yield single
            | Deep(prefix, Lazy deeper, suffix) ->
                yield! prefix |> Digit.toList
                yield! deeper |> toSeq |> Seq.collect Node.toList
                yield! suffix |> Digit.toList
            | Empty -> ()
        }

我没能编译的是这个：

module ThisDoesnt =

    module Monoids =
        type IMonoid<'m> =
            abstract Zero:'m
            abstract Plus:'m -> 'm

        type IMeasured<'m when 'm :> IMonoid<'m>> =
            abstract Measure:'m

        type Size(value) =
            new() = Size 0

            member __.Value = value

            interface IMonoid<Size> with
                member __.Zero = Size()
                member __.Plus rhs = Size(value + rhs.Value)

        type Value<'a> =
            | Value of 'a

            interface IMeasured<Size> with
                member __.Measure = Size 1

    open Monoids

    module Node =
        type Node<'m, 'a when 'm :> IMonoid<'m>> =
            | Node2 of 'm * 'a * 'a
            | Node3 of 'm * 'a * 'a * 'a

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

    module Digit =
        type Digit<'m, 'a when 'm :> IMonoid<'m>> =
            | One of 'a
            | Two of 'a * 'a
            | Three of 'a * 'a * 'a
            | Four of 'a * 'a * 'a * 'a

        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]

    module FingerTree =
        open Node
        open Digit

        type FingerTree<'m, 'a when 'm :> IMonoid<'m>> =
            | Empty
            | Single of 'a
            | Deep of 'm * Digit<'m, 'a> * Lazy<FingerTree<'m, Node<'m, 'a>>> * Digit<'m, 'a>

        let unpack (Value v) = v

        let rec toSeq<'a> (tree:FingerTree<Size, Value<'a>>) : seq<'a> = seq {
            match tree with
            | Single(Value single) ->
                yield single
            | Deep(_, prefix, Lazy deeper, suffix) ->
                yield! prefix |> Digit.toList |> List.map unpack

                #if ITERATE
                for (Value deep) in toSeq deeper do
                                    ^^^^^
                    yield deep

                #else

                yield! deeper |> toSeq |> Seq.collect (Node.toList >> List.map unpack)
                                 ^^^^^
                #endif

                yield! suffix |> Digit.toList |> List.map unpack
            | Empty -> ()
        }

我收到的错误消息说

Error Type mismatch. Expecting a
FingerTree<Size,Node<Size,Value<'a>>> -> 'b
but given a
FingerTree<Size,Value<'c>> -> seq<'c>
The type 'Node<Size,Value<'a>>' does not match the type 'Value<'b>'

并且波浪线强调了 toSeq 的递归调用。

我知道“更深层次的”类型封装在 Node 中，在工作代码中我只是在之后解压它。但是在这里，编译器在我有机会解包之前就已经出错了。尝试 for (Value deep) in toSeq deeper do yield deep 有同样的问题。

~~我已经有办法了，就是用“基”的toSeqTree，然后Seq.map unpack。~~不是是的，尝试会产生非常相似的错误消息。

我很好奇是什么导致此代码损坏以及如何修复它。

Answer 1

编译器的错误消息对我来说似乎很清楚：toSeq is applicable only to values of type FingerTree<Size, Value<'a>> for some 'a，但你试图在一个值上调用它相反，类型 FingerTree<Size,Node<Size,Value<'a>>> 是不兼容的。没有什么特定于多态递归或序列表达式，这些类型只是不匹配。

相反，通过采用 FingerTree<Size, 'a> 类型的输入（不引用 Value）使 toSeq 更通用似乎会简单得多，这将使你想要的递归调用。然后，您可以通过将更通用的 toSeq 与 Seq.map unpack.

组合起来，轻松导出您实际想要的更具体的功能

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