Overloading and virtual functions

[David Crow]

bob16972 wrote:

You don't by any chance know what is responsible, topic wise, for this behavior?

CDerivedClass does not have a two-argument DoSomething() method, hence the compiler error.

"Normal is getting dressed in clothes that you buy for work and driving through traffic in a car that you are still paying for, in order to get to the job you need to pay for the clothes and the car and the house you leave vacant all day so you can afford to live in it." - Ellen Goodman

"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne

[bob16972]

I guess I'm missing the point. I apologize if I'm missing something right under my nose. If your saying a derived class needs to implement all the virtual methods that are in the base, then I guess I'm not understanding why this works... class CBaseClass { public: CBaseClass(void) {} virtual ~CBaseClass(void) {} virtual void DoSomething(int a) {} virtual void DoSomething(int a, int b) {} }; class CDerivedClass : public CBaseClass { public: CDerivedClass(void) {} virtual ~CDerivedClass(void) {} }; int _tmain(int argc, TCHAR* argv[], TCHAR* envp[]) { CDerivedClass testing; testing.DoSomething(5,5); // Get an error on this line }

[led mike]

bob16972 wrote:

Thanks for the advice on how to cope with it. You don't by any chance know what is responsible, topic wise, for this behavior? Is this because of the concept of "Hiding"? I'm not sure what topic I need to google for to better understand it.

Polymorphism[^] Therefore, what the virtual keyword does is to allow a member of a derived class with the same name as one in the base class to be appropriately called from a pointer, and more precisely when the type of the pointer is a pointer to the base class but is pointing to an object of the derived class

led mike

[David Crow]

bob16972 wrote:

If your saying a derived class needs to implement all the virtual methods that are in the base,

They don't have to be implemented, but when they are, their types must match.

"Normal is getting dressed in clothes that you buy for work and driving through traffic in a car that you are still paying for, in order to get to the job you need to pay for the clothes and the car and the house you leave vacant all day so you can afford to live in it." - Ellen Goodman

"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne

[bob16972]

thanks. I guess the part that is confusing me is why a pointer, at any level, will invoke the base implementations when no function is overloaded (and not implemented in a derived class) but a pointer at the base class is required when functions are overloaded and an implementation for at least one overloaded function is not provided in the derived class.

[CPallini]

BTW, the following is allowed

int _tmain(int argc, TCHAR* argv[], TCHAR* envp[])
{
CBaseClass *pCDerivedClass = new CDerivedClass();
pCDerivedClass->DoSomething(5,5);
}

Please note: error check and cleanup left to the reader...

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
[my articles]

[Mark Salsbery]

hmm...what happened to inheritance? CDerivedClass is a CBaseClass. Logically, this doesn't seem like it should be an error. Until you convince me otherwise :) Mark

Mark Salsbery Microsoft MVP - Visual C++ :java:

[bob16972]

Noted. :) I had become complacent with being able to call any function implemented in the base class regardless of whether it was virtual using a pointer from any level in the derived hierarchy when there were no overloaded functions in the base class. This kinda throws me curve ball but I'm learning. Thanks for the info.

[Brian R]

You can add a using statement to promote the visibility of the base class functions. class CDerivedClass : public CBaseClass { public: CDerivedClass(void) {} using CBaseClass::DoSomething; virtual ~CDerivedClass(void) {} virtual void DoSomething(int a) {} }; The eliminates the "hiding" problem.

[bob16972]

That seems to fix the issue. :) Thanks for the help. Now I need to go back and undo the workaround code and the half baked solution I put in.