NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c++-part2/11893

Text File | 1996-08-05 | 1.7 KB | 55 lines

Path: pop.gnn.com!HoangTQ
From: Tuyen Hoang <[e-mail]>
Newsgroups: comp.lang.c++
Subject: Re: Derived class not calling overloaded base class function
Date: Sat, 16 Mar 1996 17:26:03
Organization: GNN
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an illustration
goo angoo;
angoo.foo::get(); //now the get(void) in foo gets called

If you hate this syntax -I do- and want to use angoo.get() to call the base
class function, you need to do one *easy* thing: add the follow line into
your definition of class goo

class goo : public foo{
//...
void get() { foo::get() ; }
//...
};

that's all, but the concept is not easy. If you want some explanation, you
should look back the thread and the FAQ. As my experience, I usually need an
illustration before I can understand the theorical explanation.

Then this is my question
 Do we need a more elegant way than this work around. I want to have a way to
say to the compiler that I do not declare a new function-inlining and calling
the base function- but notify that the base function is now _unhiding_ from
the derived overloaded function, something look like

class derived : public base{
//....
    public:
       void get(int);
  /* unhiding */ void base::get();
//....
};

the problem is not concern about one or more overhead calls but the concept
that the base function is now *explicit* present in the derived class and can
be overloaded as it was declared here. Does the virtual key word is
needed/no needed?

Regard
Tuyen Hoang