Expressing Designs in C# - Effective C# - page 140

Notice that I had to add the new keyword to the definition of the previous
Message method (see Item 33). MyClass.Message() is not virtual. Derived
classes cannot provide an overridden version of Message. The MyDerived
class creates a new Message method, but that method does not override
MyClass.Message: It hides it. Furthermore, MyClass.Message is still avail-
able through the IMsg reference:
MyDerivedClass d = new MyDerivedClass ();
d.Message(); // prints "MyDerivedClass".
IMsg m = d as IMsg ;
m.Message(); // prints "MyClass"
Interface methods are not virtual. When you implement an interface, you
are declaring a concrete implementation of a particular contract in that
type.
But you often want to create interfaces, implement them in base classes,
and modify the behavior in derived classes. You can. You've got two
options. If you do not have access to the base class, you can reimplement
the interface in the derived class:
public class MyDerivedClass : MyClass
{
public new void Message()
{
Console .WriteLine( "MyDerivedClass" );
}
}
The addition of the IMsg keyword changes the behavior of your derived
class so that IMsg.Message() now uses the derived class version:
MyDerivedClass d = new MyDerivedClass ();
d.Message(); // prints "MyDerivedClass".
IMsg m = d as IMsg ;
m.Message(); // prints " MyDerivedClass "
Yo u s t i l l n e e d t h e n e w k e y w o r d o n t h e M y D e r i v e d C l a s s . M e s s a g e ( ) m e t h o d .
That's your clue that there are still problems (see Item 33). The base class
version is still accessible through a reference to the base class:
MyDerivedClass d = new MyDerivedClass ();
d.Message(); // prints "MyDerivedClass".