Please help understand virtual functions

[cctv.star]

I have a base class that implements IDisposable.

Then I have a derived class, in which I write the following:

public override void Dispose() {
}

... only to get an error: cannot override inherited member ...Dispose()
because it is not marked virtual, abstract, or override. Why??? How a
function, implementing an interface, not be virtual?

And how can I do what I want?

[Mike Urquiola]

Just mark the base method as virtual

public class MyClass:IDisposable {

#region IDisposable Members

public virtual void Dispose() {

throw new NotImplementedException();

}

#endregion

}

public class MyClass2 : MyClass {

#region IDisposable Members

public override void Dispose() {

base.Dispose();

throw new NotImplementedException();

}

#endregion

}

<cc*******@gmail.comwrote in message
news:55**********************************@t54g2000 hsg.googlegroups.com...

>I have a base class that implements IDisposable.

Then I have a derived class, in which I write the following:

public override void Dispose() {
}

.. only to get an error: cannot override inherited member ...Dispose()
because it is not marked virtual, abstract, or override. Why??? How a
function, implementing an interface, not be virtual?

And how can I do what I want?

[cctv.star]

Doesn't declaring a function a member of an interface imply that it's
virtual?

[Rudy Velthuis]

cc*******@gmail.com wrote:

Doesn't declaring a function a member of an interface imply that it's
virtual?

No.

--
Rudy Velthuis http://rvelthuis.de

"In this war – as in others – I am less interested in honoring
the dead than in preventing the dead." -- Butler Shaffer

[Marc Gravell]

No.

Well, that's a bit of an over-simplification; from the implementing
class's perspective (the callee) - then indeed it isn't automatically
virtual, and you can't override etc; but from the caller's
perspective, it is a virtcall, since it can't possibly static-bind to
an interface method, and the JIT won't ever attempt to inline etc.

Marc

[Rudy Velthuis]

Marc Gravell wrote:

No.

Well, that's a bit of an over-simplification; from the implementing
class's perspective (the callee) - then indeed it isn't automatically
virtual, and you can't override etc; but from the caller's
perspective, it is a virtcall, since it can't possibly static-bind to
an interface method, and the JIT won't ever attempt to inline etc.

Sure, calling an interface method is similar to calling a virtual
method (the internal mechanism is very similar, indeed). But the method
is not necessarily virtual in the implementing class.
--
Rudy Velthuis http://rvelthuis.de

"All truth passes through three stages. First, it is ridiculed.
Second, it is violently opposed. Third, it is accepted as being
self-evident."
-- Arthur Schopenhauer (1788-1860)

[Charles Calvert]

On Wed, 23 Apr 2008 12:47:07 -0700 (PDT), cc*******@gmail.com wrote in
<55**********************************@t54g2000hsg. googlegroups.com>:

>I have a base class that implements IDisposable.

Then I have a derived class, in which I write the following:

public override void Dispose() {
}

.. only to get an error: cannot override inherited member ...Dispose()
because it is not marked virtual, abstract, or override. Why?

Are you asking why you got the error?

How a function, implementing an interface, not be virtual?

An interface is a list of methods and properties that you must
implement. You're not inheriting from a class, just making your class
conform to a standard. There is no base implementation of Dispose to
override.

>And how can I do what I want?

Look at the docs for IDisposable.Dispose. There is plenty of
information and an example.
--
Charles Calvert | Software Design/Development
Celtic Wolf, Inc. | Project Management
http://www.celticwolf.com/ | Technical Writing
[phone] | Research

[Ben Voigt [C++ MVP]]

cc*******@gmail.com wrote:

I have a base class that implements IDisposable.

Then I have a derived class, in which I write the following:

public override void Dispose() {
}

.. only to get an error: cannot override inherited member ...Dispose()
because it is not marked virtual, abstract, or override. Why??? How a
function, implementing an interface, not be virtual?

And how can I do what I want?

You could also implement IDisposable in the derived class, this
implementation will override the base class. However if the base class did
an explicit override, you will not be able to invoke it from your derived
class.