interface - 定义一个继承接口(interface)但不实现接口(interface)的抽象类

Question

合并它编译的 leppies 反馈 - 但 IMO 有一些缺点，我希望编译器强制每个子类定义它们自己的 Uri 属性。现在的代码:

[<AbstractClass>] 
type UriUserControl() = 
    inherit UserControl()
    interface IUriProvider with 
        member this.Uri with get() = null

有趣的是，我定义的类从上面定义了哪些 inerits 不显示公共(public) Uri 属性:

type Page2() as this =
    inherit UriUserControl()
    let uriStr = "/FSSilverlightApp;component/Page2.xaml"
    let mutable uri = new System.Uri(uriStr, System.UriKind.Relative)
    do
        Application.LoadComponent(this, uri)

    member public this.Uri with get () = uri

我想定义一个继承自 UserControl 和我自己的接口(interface) IUriProvider 的抽象类，但没有实现它。目标是能够定义实现 UserControl 但也提供自己的 Uri 的页面(用于 silverlight)(然后将它们粘贴在列表/数组中并将它们作为一组处理:

type IUriProvider = 
interface
    abstract member uriString: String ;
    abstract member Uri : unit -> System.Uri ;
end

[<AbstractClass>] 
type UriUserControl() as this = 
    inherit IUriProvider with
        abstract member uriString: String ;
    inherit UserControl()

还有定义中的 Uri - 我想实现为属性 getter - 并且也遇到了问题。

这不编译

type IUriProvider = 
    interface
        abstract member uriString: String with get;
end

Answer 1

这是一种方法:

type IUriProvider =  
    abstract member UriString: string
    abstract member Uri : System.Uri

[<AbstractClass>]  
type UriUserControl() as this =  
    inherit System.Windows.Controls.UserControl() 
    abstract member Uri : System.Uri
    abstract member UriString : string
    interface IUriProvider with 
        member x.Uri = this.Uri
        member x.UriString = this.UriString

请注意，您必须提供接口(interface)的实现(因为 F# 中的所有接口(interface)实现都是显式的)，但这只能引用类中的抽象成员。然后你可以这样子类化:

type ConcreteUriUserControl() =
    inherit UriUserControl()
    override this.Uri = null
    override this.UriString = "foo"