types - 不可能用递归类型约束创建对象？

Question

这是一个纯粹的学术问题，但来自 this question
关于类型约束。提问者举了一个例子:

type Something<'a, 'b when 'b :> seq<'b>>() =
    member __.x = 42

f# 愉快地编译。现在的问题是你如何制作这个对象？

let z = new Something<???, ???>()

Answer 1

type T() =
    interface seq<T> with
        member this.GetEnumerator() = ([] :> seq<T>).GetEnumerator()
        member this.GetEnumerator() = ([] :> seq<T>).GetEnumerator() :> System.Collections.IEnumerator

let z = new Something<string, T>()