Question

给出以下界面：

type A<'t> =
    abstract Scratch: 't

如何简单地创建这样的界面：

type X<A<'t>> =
    abstract f: A<'t>

我也试过这个：

type X<'t,'y when 'y:A<'t>> =
    abstract f: 'y

和此：

type X<'t,A<'t>> =
    abstract f: A<'t>

但没有一个适合我。

Answer 1

如果你想以界面的方式试试这个：

type IA<'a> =
    abstract Scratch: 'a

type IX<'a, 'b when 'a :> IA<'b>> = 
    abstract F: 'a

type RealA() = 
    interface IA<int> with
        member __.Scratch = 1

type RealX = 
    interface IX<RealA, int> with
        member __.F = RealA()

如果你想要抽象类：

[<AbstractClass>]
type A<'a>() = 
    abstract member Scratch: 'a

[<AbstractClass>]
type X<'a, 'b when 'a :> A<'b>>() = 
    abstract F: 'a

type RealA'() = 
    inherit A<int>()
    override __.Scratch = 1

type RealX'() = 
    inherit X<RealA', int>()
    override __.F = RealA'()

嵌套通用类型的接口

1 个答案: