带运算符的协变泛型？

Question

Is there any way to have this class's generic argument be covariant while keeping the operator?

public class ContentReference<T> where T : IReferable
{

    public string Name { get; private set; }

    public ContentReference(T value)
    {
        this.Name = value.Name;
    }        

    public static implicit operator ContentReference<T>(T value)
    {
        return new ContentReference<T>(value);
    }

}

So that I can have, say, a ContentReference<Audio> be assigned to by a ContentReference<SoundEffect> or ContentReference<MusicTrack> ?

Answer 1

The response is no... What you can do:

public abstract class ContentReference
{
    public string Name { get; private set; }

    protected ContentReference(string name)
    {
        Name = name;
    }

    public abstract void Play();
}

public class ContentReference<T> where T : IReferable
{
    public ContentReference(T value) : base(value.Name)
    {
    }        

    public static implicit operator ContentReference<T>(T value)
    {
        return new ContentReference<T>(value);
    }

    public override void Play()
    {
        // Play the IReference 
    }
}

So the solution is that you can have a base class (optionally abstract ) that exposes the common methods that are partially implemented in the base class and partially implemented in the subclass(es).

Or you could define a contravariant IContentReference<out T> :

public interface IContentReference<out T> where T : IReferable
{
    string Name { get; }
}

public class ContentReference<T> : IContentReference<T> where T : IReferable
{
    public string Name { get; private set; }

    public ContentReference(T value)
    {
        this.Name = value.Name;
    }

    public static implicit operator ContentReference<T>(T value)
    {
        return new ContentReference<T>(value);
    }
}

and then:

IContentReference<Audio> interf1 = new ContentReference<SoundEffect>(new SoundEffect());
IContentReference<Audio> interf2 = new ContentReference<MusicTrack>(new MusicTrack());

IContentReference<Audio> interf3 = (ContentReference<SoundEffect>)new SoundEffect();
IContentReference<Audio> interf4 = (ContentReference<MusicTrack>)(new MusicTrack());