f# - 移动递归类型时代码中断

Question

在重构一些代码的过程中，我注意到一种情况，即代码在移动时会中断:

type ThingBase () = class end

and Functions =
    static member Create<'T when 'T :> ThingBase and 'T : (new : Unit -> 'T)> () = new 'T ()

and Thing () =
    inherit ThingBase ()
    static member Create () = Functions.Create<Thing> ()

// This works, but try moving the Functions type here instead.

如果将 Functions 类型移动到 Thing 类型下方，代码将意外中断:

type ThingBase () = class end

and Thing () =
    inherit ThingBase ()
    static member Create () = Functions.Create<Thing> ()
//                            ^^^^^^^^^^^^^^^^^^^^^^^^^^
// This construct causes code to be less generic than indicated by the
// type annotations. The type variable 'T has been constrained to be
// type 'Thing'.

and Functions =
    static member Create<'T when 'T :> ThingBase and 'T : (new : Unit -> 'T)> () = new 'T ()
//                ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// The type ''T' does not match the type 'Thing'.

而且无论我尝试什么，我都无法对其进行类型检查。为什么类型推断如此顽固，拒绝泛化 Create 方法。

顺便说一句，我还尝试了 F# 4.1 module rec 并且 Create 是否是模块中的函数也没有关系。

有什么见解吗？在我看来，编译器应该不会遇到任何问题。

Answer 1

如果你这样做，它会编译

static member Create<'T when 'T :> ThingBase 
         and 'T : (new : Unit -> 'T)> () : 'T = new 'T ()
//                                       ^^^^          

明确说明返回类型的地方。

递归定义分两次从左到右进行类型检查；首先是函数/方法签名，然后是主体。您需要正文(或显式返回类型注释)来获取结果类型，因此您要么首先需要正文，要么需要注释，以便它在两遍中的第一遍中得到解决。