如何使Abstract类继承F＃中的接口？

Question

我有带有接口和抽象类的以下代码。

    type INode<'a> =
    abstract member Parent:INode<'a>
    abstract member Evaluate:unit->'a
    abstract member Clone:INode<'a> -> INode<'a>
    abstract member SelectRandom:int->INode<'a> option

[<AbstractClass>]
type TerminalNode<'a>() =
    interface INode<'a> with
        member this.SelectRandom indexOfCurrentNode = 
            let p = 1.0 / float indexOfCurrentNode
            if Random().NextDouble() < p then Some (this:>INode<'a>) else None

我想在其中实现层次结构 但是编译器抱怨应该实现该接口的所有方法。

创建实现一个接口并实现该接口的一个方法而使其他方法保持抽象的Abstract类的正确方法是什么？

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Answer 1

您可以在类型上定义抽象成员，然后提供指向这些抽象成员的实现。例如：

[<AbstractClass>]
type TerminalNode<'a>() as self =
    abstract member Parent : INode<'a>
    abstract member Evaluate : unit->'a
    abstract member Clone:INode<'a> -> INode<'a>
    interface INode<'a> with
        member this.Parent = self.Parent
        member this.Evaluate () = self.Evaluate ()
        member this.Clone a = self.Clone a
        member this.SelectRandom indexOfCurrentNode = 
            ...

（注意，as self在type线路，它可以让你通过变量名指的是抽象类self）