Solved

Why doesn't the compiler see overloaded base class functions?

Posted on 2002-05-15
7
198 Views
Last Modified: 2010-04-02
Why doesn't the compiler see overloaded base class functions? I cannot explain why the follwing is an error:

    #include <string>
    #include <iostream>
    using namespace std;

    class Base
    {
      public:
        virtual void Do( const string& )
        {
          cout << "Base::Do( string )" << endl;
        }
        virtual void Do( const char* const data,
                                  const long length )
        {
          cout << "Base::Do( char*, long )" << endl;
        }
    };

    class Derived : public Base
    {
      public:
        virtual void Do( const int s )
        {
          cout << "Base::Do( int )" << endl;
        }
    };

    int main(int argc, char* argv[])
    {
      Derived d;
      d.Do( 5 );
      d.Do( string() );
      d.Do( "foobar", 6 );
      return 0;
    }

Why doesn't the compiler "see" the base class versions of Do()? The only way I can get this to compile is to qualify the calls by adding "Base::", like so:

    d.Base::Do( string() );
    d.Base::Do( "foobar", 6 );

Can someone explain to me why this is the case? Thanks!
0
Comment
Question by:magenta
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
7 Comments
 

Author Comment

by:magenta
ID: 7012178
I should have pointed out that the compiler complains about the following lines:

     d.Do( string() );
     d.Do( "foobar", 6 );

It seems to only see the Derived methods.
0
 
LVL 6

Accepted Solution

by:
thienpnguyen earned 50 total points
ID: 7012402
In subclass, if you have a overload funtion that is same name with a fucntion in base class, then that function in base class will be hide .

0
 
LVL 6

Expert Comment

by:thienpnguyen
ID: 7012417
For fixing the bug, you can do as following

class Derived : public Base
{
public:

    using  Base::Do; // <-------- new code

    virtual void Do( const int s )
    {
      cout << "Base::Do( int )" << endl;
    }

};

int main(int argc, char* argv[])
{
  Derived d;
  d.Do( 5 );
  d.Do( string() );
  d.Do( "foobar", 6 );
  return 0;
}
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 6

Expert Comment

by:thienpnguyen
ID: 7012427
0
 
LVL 3

Expert Comment

by:MDarling
ID: 7012450
This is not meant to trample on thienpnguyen's answer.

Just thought I'd post the full listing as I got it to compile on M$ VC6.0.

Nice links Thien! Almost perfect for this question.




#include <string>
#include <iostream>
using namespace std;

class Base
{
public:
     virtual void Do( const string& )
     {
          cout << "Base::Do( string )" << endl;
     }
     virtual void Do( const char* const data,
          const long length )
     {
          cout << "Base::Do( char*, long )" << endl;
     }
};

class Derived : public Base
{
public:
     
     Base::Do;
     
     virtual void Do( const int s )
     {
          cout << "Base::Do( int )" << endl;
     }
};

int main(int argc, char* argv[])
{
     Derived d;
     d.Do( 5 );
     d.Do( string() );
     d.Do( "foobar", 6 );
     return 0;
}
0
 

Author Comment

by:magenta
ID: 7012541
Very intesting...I'm going to have to sit down and read that article. I should probably have the C++ standard nearby...

However, before I do that, answer this question. If a derived class by definition can be treated the same as a base class, then isn't the code above example of this rule being broken?
0
 
LVL 11

Expert Comment

by:griessh
ID: 7178762
Dear magenta

I think you forgot this question. I will ask Community Support to close it unless you finalize it within 7 days. You can always request to keep this question open. But remember, experts can only help you if you provide feedback to their questions.
Unless there is objection or further activity,  I will suggest to accept

     "thienpnguyen"

comment(s) as an answer.
     "refund the points and delete this question"
since nobody had a satisfying answer for you.
since you never gave more feedback.
PAQ at zero points.

If you think your question was not answered at all, you can post a request in Community support (please include this link) to refund your points. The link to the Community Support area is: http://www.experts-exchange.com/commspt/


PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER!
======
Werner
0

Featured Post

Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

What is C++ STL?: STL stands for Standard Template Library and is a part of standard C++ libraries. It contains many useful data structures (containers) and algorithms, which can spare you a lot of the time. Today we will look at the STL Vector. …
This article shows you how to optimize memory allocations in C++ using placement new. Applicable especially to usecases dealing with creation of large number of objects. A brief on problem: Lets take example problem for simplicity: - I have a G…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.
The viewer will be introduced to the member functions push_back and pop_back of the vector class. The video will teach the difference between the two as well as how to use each one along with its functionality.

726 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question