为什么此类型不是接口？

Question

我想为默认函数实现一个相等和比较功能的接口。

如果我从成员中删除IKeyable<'A>除Key成员之外的所有内容，则它是有效接口，只要不添加默认实现即可。从中删除其他接口实现IKeyable<'A>，并仅保留默认成员将以相同的结果结束。

type IKeyable<'A when 'A: equality and 'A :> IComparable> = 
    abstract member Key : 'A

    default this.Equals obj = // hidden for clarity

    default this.GetHashCode () = // hidden for clarity

    interface IEquatable<'A> with
        member this.Equals otherKey = // hidden for clarity

    interface IComparable<'A> with
        member this.CompareTo otherKey = // hidden for clarity

    interface IComparable with
        member this.CompareTo obj = // hidden for clarity

type Operation = 
    { Id: Guid }
    interface IKeyable<Guid> with // Error: The type 'IKeyable<Guid>' is not an interface type
        member this.Key = this.Id

我想将其IKeyable<'A>用作接口，以便“获得”用于相等性和比较的默认实现。

错误消息出现interface ... with在类型下 Operation：The type 'IKeyable<Guid>' is not an interface type

Answer 1

接口不能有方法的实现，和你的类型有其中五- ，Equals，GetHashCode，IEquatable<_>.Equals，IComparable<_>.CompareTo和IComparable.CompareTo。

接口纯粹是一组方法和属性。它不像基类，它不能给实现者一些“默认”实现，基本行为或实用程序方法。

要使您的类型成为接口，请摆脱所有实现：

type IKeyable<'A when 'A: equality and 'A :> IComparable> = 
    inherit IEquatable<'A>
    inherit IComparable<'A>
    abstract member Key : 'A

如果您确实想保留默认的实现，则必须使其成为基类而不是接口，在这种情况下，Operation必须将其变为类而不是记录：

type Operation(id: Guid)
    inherit IKeyable<Guid>
    override this.Key = id
    member val Id = id