Type parameter 'T' has the same name as the type parameter from outer type '…'

Question

public abstract class EntityBase { ... }

public interface IFoobar
{
    void Foo(int x)
        where T : EntityBase, new();
}

public interface IFoobar<         


        

Answer 1

You can do one of two things:

1. Ignore the warning and make both types T.

2. Do a run-time check and throw an exception:

   if (typeof(T) != typeof(U)) throw Exception("Not the same type");
   

As others have stated, perhaps you need to rethink the way you are designing your interfaces.