C#中的嵌套接口

Question

我正在尝试实现一个IFunctor接口[Haskell的Fuctor类]和一个Maybe创建两个类的接口:Just和Nothing.

到目前为止,我有:

public interface IFunctor<A> {
    IFunctor<B> fmap<B> (Func<A,B> f);
}

public interface Maybe<A> : IFunctor<A> {
    Maybe<B> fmap<B> (Func<A,B> f);
}

public class Nothing<A> : Maybe<A> {

    public Maybe<B> fmap<B>( Func<A,B> f ){
        return new Nothing<B>();
    }
}

但是我得到了

`Pr.Nothing<A>' does not implement interface member `IFunctor<A>.fmap<B>
(System.Func<A,B>)' and the best implementing candidate `Pr.Nothing<A>.fmap<B>
(System.Func<A,B>)' return type `Pr.Maybe<B>' does not match interface member return
type `IFunctor<B>'

不是Maybe<B>会员IFunctor<B>？

解

我结束了写作

public interface IFunctor<A> {
    IFunctor<B> fmap<B> (Func<A,B> f);
}

public interface Maybe<A> : IFunctor<A> {
    //Some stuff but not fmap
}

public class Nothing<A> : Maybe<A> {

    public IFunctor<B> fmap<B>( Func<A,B> f ){
        return new Nothing<B>();
    }
}

Answer 1

Maybe<A>.fmap()没有override IFunctor<A>.fmap().任何类型的实现Maybe<A>需要需要实现两个 Maybe<A>和IFunctor<A>.

public interface IFunctor<A>
{
    IFunctor<B> fmap<B>(Func<A, B> f);
}

public interface Maybe<A> : IFunctor<A>
{
    Maybe<B> fmap<B>(Func<A, B> f);
}

public class Nothing<A> : Maybe<A>
{

    public Maybe<B> fmap<B>(Func<A, B> f)
    {
        return new Nothing<B>();
    }

    //This is the explicit implementation of IFunctor<A>.fmap<B>
    //which in turn invokes method above.
    IFunctor<B> IFunctor<A>.fmap<B>(Func<A, B> f)
    {
        return this.fmap(f);
    }
}