Why should Dispose() be non-virtual?

Question

I\'m new to C#, so apologies if this is an obvious question.

In the MSDN Dispose example, the Dispose method they define is non-virtual. Why is that? It seems odd

Answer 1

Although methods in an interface are not "virtual" in the usual sense, they can nevertheless still be implemented in classes that inherit them. This is apparently a convenience built into the C# language, allowing the creation of interface methods without requiring the virtual keyword, and implementing methods without requiring the override keyword.

Consequently, although the IDisposable interface contains a Dispose() method, it does not have the virtual keyword in front of it, nor do you have to use the override keyword in the inheriting class to implement it.

The usual Dispose pattern is to implement Dispose in your own class, and then call Dispose in the base class so that it can release the resources it owns, and so on.

A type's Dispose method should release all the resources that it owns. It should also release all resources owned by its base types by calling its parent type's Dispose method. The parent type's Dispose method should release all resources that it owns and in turn call its parent type's Dispose method, propagating this pattern through the hierarchy of base types.

http://msdn.microsoft.com/en-us/library/fs2xkftw.aspx