Re: Can a virtual method be overloaded ?

"Victor Bazarov" <>
Mon, 24 Apr 2006 22:31:19 -0400
Bit byte wrote:

Victor Bazarov wrote:

Bit byte wrote:

I have a class BaseApplication, from which I am deriving two
classes A and B.

In class BaseApplication, I have a virtual method as ff:

virtual void saveToDb(const char*);

So, you essentially have

    class BaseApplication
        virtual void saveToDb(const char*);

, right? This is much easier to understand.


I have this :

    class BaseApplication
        virtual void saveToDb(const char*);

   class A: public BaseApplication {

void saveToDb( const struct_for_a*) ;

This member does not override the 'saveToDb' you declared in the base
class. It _hides_ it. If you intended to override the function, you
need to keep its signature the same, IOW, the argument has to be the
same -- const char*. As soon as you changed it, the function is not
the overrider any longer.


    //idea behind this is that B is a type of application that will
    //need to persist many different types of data (polymorphically)
   class B: public BaseApplication {

void saveToDb( const struct relevant_for_b*) ;
void saveToDb( const struct another_struct_for_b*) ;
void saveToDb( const struct yet_another_for_b*) ;

 I want to know if I can do this. I believe I have done something
similar to this many moons ago, but I don't quite remember.

You can do this. It has nothing to do with polymorphism, but it is
perfectly legal.

   #include <iostream>
   struct A {
       virtual void foo(const char*) { std::cout << "A::foo()\n"; }

   struct B : A {
       void foo(int) { std::cout << "B::foo()\n"; }

   struct OK : A {
       void foo(const char*) { std::cout << "OK::foo()\n"; }

   int main() {
      B b;
      OK ok;

      A *pa = &b;
      pa->foo("abc"); // base class function is called

      pa = &ok;
      pa->foo("def"); // derived class function is called

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