Short answer: It is not possible. See this post
An Interface can't derive from a class
. However nothing prevent you from doing this:
public interface ISomeView
{
}
public interface IView<out T> where T:class
{
T SomeParam { get; }
}
public class SomeView:IView<ISomeView>
{
public ISomeView SomeParam { get; set; }
}
Edit:
If you don't want to implement the T SomeParam { get; }
each time you need to have an implementation, Does this would work?
public interface ISomeView
{
}
public abstract class BaseView<T> where T : class
{
public T SomeParam { get; set; }
}
public class SomeView : BaseView<ISomeView>{
}
In both case this would work:
public class main
{
public class OneOfThoseView : ISomeView
{
}
public main()
{
OneOfThoseView oneOfThose = new OneOfThoseView();
SomeView x = new SomeView();
x.SomeParam = oneOfThose;
}
}
Edit 2:
Not exactly what you want to do but this would force your SomeView class to return a BaseView<SomeView>
class
public interface ISomeView
{
}
public abstract class BaseView<T> where T : BaseView<T>
{
public T SomeParam { get; set; }
}
public class SomeView : BaseView<SomeView>
{
}
Now only this would work.
public main()
{
SomeView y= new SomeView ();
SomeView x = new SomeView();
x.SomeParam = y;
}