C#: Property overriding by specifying the interface explicitly

Question

While attempting to override the explicit interface implementation of the ICollection<T>.IsReadOnly property from the Collection<T> class, I came across some documents stating that explicit interface member implementations cannot be overridden because they cannot have modifiers such as virtual or abstract. On MSDN they even go as far as specifying how to make an explicit interface member implementation available for inheritance by creating another abstract or virtual member which is called by the explicit interface member implementation. No problems so far.

But then I wonder: Why is it possible in C# to override any explicitly implemented interface member just by specifying the interface explicitly?

For example, suppose I have a simple interface like this, with a property and method:

public interface IMyInterface
{
    bool AlwaysFalse { get; }
    bool IsTrue(bool value);
}

And a class A which implements the interface explicitly, and has a method Test() which calls its own interface member implementation.

public class A : IMyInterface
{
    bool IMyInterface.AlwaysFalse
    { get { return false; } }

    bool IMyInterface.IsTrue(bool value)
    { return value; }

    public bool Test()
    { return ((IMyInterface)this).AlwaysFalse; }
}

As you can see, none of the four members are virtual or abstract, so when I define a class B like this:

public class B : A
{
    public bool AlwaysFalse
    { get { return true; } }

    public bool IsTrue(bool value)
    { return !value; }
}

Then you'd expect an instance of B cast to A to behave like A. And it does:

A a = new A();
Console.WriteLine(((IMyInterface)a).AlwaysFalse);    // False
Console.WriteLine(((IMyInterface)a).IsTrue(false));  // False
Console.WriteLine(a.Test());                         // False
A b = new B();
Console.WriteLine(((IMyInterface)b).AlwaysFalse);    // False
Console.WriteLine(((IMyInterface)b).IsTrue(false));  // False
Console.WriteLine(b.Test());                         // False

Now comes the catch. Create a class C which is an exact copy of B except for one thing in the class declaration:

public class C : A, IMyInterface
{ /* ... same as B ... */ }

Now an instance of C, when cast to A, doesn't behave like A but like C:

A c = new C();
Console.WriteLine(((IMyInterface)c).AlwaysFalse);    // True
Console.WriteLine(((IMyInterface)c).IsTrue(false));  // True
Console.WriteLine(c.Test());                         // True

Even the Test() method now calls the overridden method in C! Why is this?

Answer 1

When you implement an interface explicitly , as in class c and you create the object of that class and cast it to the implemented interface then that instance will work on the implemented interface.

And it sounds logical also because If a class is implementing an interface so that interface and work.

Answer 2

This has nothing to do with explicit interface implementation; it's simply a consequence of the general rules of inheritance and interface mapping: you would see exactly the same results if type A provided an implicit, rather than explicit, implementation of IMyInterface.

• Type B inherits from type A. Nothing is overridden.
  B provides its own AlwaysFalse and IsTrue members but they don't implement IMyInterface; The implementation of IMyInterface is provided by the members inherited from A: when an instance of type B is cast to IMyInterface then it behaves in exactly the same way as an instance of type A because A is providing the members that implement the interface.
• Type C inherits from type A. Again, nothing is overridden.
  C provides its own AlwaysFalse and IsTrue members but this time those members do implement IMyInterface: when an instance of type C is cast to IMyInterface then the members of C provide the interface implementation rather than those of A.

Because type A implements IMyInterface explicitly the compiler doesn't warn that the members of B and C are hiding the members of A; In effect those members of A were already hidden due to the explicit interface implementation.

If you changed type A to implement IMyInterface implicitly rather than explicitly then the compiler would warn that the members of B and C are hiding, not overriding, the members of A and that you should ideally use the new modifier when declaring those members in B and C.

Here are some relevant bits from the language specification. (Sections 20.4.2 and 20.4.4 in the ECMA-334 spec; sections 13.4.4 and 13.4.6 in the Microsoft C#4 spec.)

20.4.2 Interface mapping

The implementation of a particular interface member I.M, where I is the interface in which the member M is declared, is determined by examining each class or struct S, starting with C and repeating for each successive base class of C, until a match is located.

20.4.4 Interface re-implementation

A class that inherits an interface implementation is permitted to re-implement the interface by including it in the base class list. A re-implementation of an interface follows exactly the same interface mapping rules as an initial implementation of an interface. Thus, the inherited interface mapping has no effect whatsoever on the interface mapping established for the re-implementation of the interface.