Should be a newbie question...
I have existing code in an existing class, A, that I want to extend in order to override an existing method, A::f().
So now I want to create class B to override f(), since I don't want to just change A::f() because other code depends on it.
To do this, I need to change A::f() to a virtual method, I believe.
My question is besides allowing a method to be dynamically invoked (to use B's implementation and not A's) are there any other implications to making a method virtual? Am I breaking some kind of good programming practice? Will this affect any other code trying to use A::f()?
Please let me know.
Thanks, jbu
edit: my question was more along the lines of is there anything wrong with making someone else's method virtual? even though you're not changing someone else's implementation, you're still having to go into someone's existing code and make changes to the declaration.
If you make the function virtual inside of the base class, anything that derives from it will also have it virtual.
Once virtual, if you create an instance of A, then it will still call A::f.
If you create an instance of B and store it in a pointer of type A*. And then you call A*::->f, then it will call B's B::f.
As for side effects, there probably won't be any side effects, other than a slight (unnoticeable) performance loss.
There is a very small side effect as well, there could be a class C that also derives from A, and it may implement C::f, and expect that if A*::->f was called, then it expects A::f to be called. But this is not very common.
But more than likely, if C exists, then it does not implement C::f at all, and in which case everything is fine.
Be careful though, if you are using an already compiled library and you are modifying it's header files, what you are expecting to work probably will not. You will need to recompile the header and source files.
You could consider doing the following to avoid side effects:
A2 that derives from A and make it's f virtual
A2 instead of AB from type A2. Depending on what you need you may also be able to use a has-a relationship instead of a is-a.
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