F#中的显式类型递归 [英] Explicit type recursion in F#

问题描述

灵感来自这个问题:

在F#中是否可以进行显式类型递归?

type 'a Mu = In of 'a Mu 'a

let unIn (In x) = x

不幸的是，这段代码给出了类型参数不能用作类型构造函数.

This code unfortunatly gives "Type parameter cannot be used as type constructor.

备注:本文函数式编程中使用了此构造例如，具有重载和高阶多态性.

用法示例(摘自此处):

type ('a, 'b) ListX =
    | Nil
    | Cons of 'a * 'b

type 'a List = ListX Mu

推荐答案

否，这是不可能的.具体而言，F#中的泛型与CLR具有相同的限制，即< T>或<'a>必须具有类型"*".同样的限制意味着您无法直接在F#中编写类型类"，因为例如"Monad m"将采用种类较高的参数"m"(例如"*-> *"，其中"list"和"option"可以是实例，它们各自都是泛型类型构造函数)，但是不允许这样做

No, this is not possible. Specifically, generics in F# have the same limitation as the CLR, namely a <T> or an <'a> must have kind " * ". This same limitation is what means you cannot author "type classes" directly in F#, since e.g. "Monad m" would take a higher-kinded argument 'm' (e.g. "* -> *", where e.g. 'list' and 'option' could be instances, those each themselves being generic type constructors), but this is not allowed.

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