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嵌套的通用接口类型

nested-generics f# generics

3

Mohsen · 技术社区 · 8 年前

给定以下接口:

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

如何简单地创建这样的接口:

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

还有这个:

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

但对我来说没用。

1 回复 | 直到 8 年前

1

5

Szer 8 年前

如果要将其作为接口执行,请尝试以下操作:

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()

如果需要抽象类:

[<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'()