无限类型(又名递归类型)在 F# 中是不可能的吗? [英] Are Infinite Types (aka Recursive Types) not possible in F#?

问题描述

我在 twitter 上与 Sadek Drobi 聊天时被提到 F# 似乎不支持无限类型.事实证明，在 C# 中，您可以按照以下方式做一些事情:

delegate RecDelegateRecDelegate (T x)；

但是，经过我们双方的一些实验后，我们确定 F# 中的相同似乎不可能隐式和显式.

显式:

type 'a specialF = 'a->specialF<'a>

<块引用>

错误 FS0191:此类型定义涉及立即循环引用通过缩写，struct 字段或继承关系.

隐式:

let rec specialF (x: 'a) = specialF

<块引用>

类型不匹配.期待一个 'b 但是给定一个'a -> 'b.结果类型统一''b'时将是无限的和 ''a -> 'b'.

当然，这些都是有意为之的简单示例.

我想知道我是不是弄错了.也许我错过了某种必要的注释?

解决方案

你也可以这样做

type 'a RecType = RecType of ('a -> 'a RecType)

创建一个命名类型，通过它执行递归.现在这有效:

let rec specialF = RecType (fun _ -> specialF)

I was chatting with Sadek Drobi on twitter when be brought up that F# didn't seem to support Infinite Types. It turns out that in C# you can do something along these lines:

delegate RecDelegate<T> RecDelegate<T>(T x);

However, after some experimentation on both our parts, we determined that the same in F# seems impossible both implicit and explicitly.

Explicit:

type 'a specialF = 'a->specialF<'a>

error FS0191: This type definition involves an immediate cyclic reference through an abbreviation, struct field or inheritance relation.

Implicit:

let rec specialF (x: 'a) = specialF

Type mismatch. Expecting a 'b but given a 'a -> 'b. The resulting type would be infinite when unifying ''b' and ''a -> 'b'.

Of course, these are intentionally simple samples.

I was wondering if I am somehow mistaken. Perhaps I missed some type of necessary annotation?

解决方案

You can also do something like

type 'a RecType = RecType of ('a -> 'a RecType)

to create a named type through which to perform the recursion. Now this works:

let rec specialF = RecType (fun _ -> specialF)

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