Calling a Superclass Virtual Function

Problem

You need to invoke a function on a superclass of a particular class, but it is overridden in subclasses, so the usual syntax of p->method( ) won't give you the results you are after.

Solution

Qualify the name of the member function you want to call with the target base class; for example, if you have two classes. (See Example 8-16.)

Example 8-16. Calling a specific version of a virtual function

#include 

using namespace std;

class Base {
public:
 virtual void foo( ) {cout << "Base::foo( )" << endl;}
};

class Derived : public Base {
public:
 virtual void foo( ) {cout << "Derived::foo( )" << endl;}
};

int main( ) {
 Derived* p = new Derived( );

 p->foo( ); // Calls the derived version
 p->Base::foo( ); // Calls the base version
}

 

Discussion

Making a regular practice of overriding C++'s polymorphic facilities is not a good idea, but there are times when you have to do it. As with so many techniques in C++, it is largely a matter of syntax. When you want to call a specific base class's version of a virtual function, just qualify it with the name of the class you are after, as I did in Example 8-16:

p->Base::foo( );

This will call the version of foo defined for Base, and not the one defined for whatever subclass of Base p points to.

Building C++ Applications

Code Organization

Numbers

Strings and Text

Dates and Times

Managing Data with Containers

Algorithms

Classes

Exceptions and Safety

Streams and Files

Science and Mathematics

Multithreading

Internationalization

XML

Miscellaneous

Index



C++ Cookbook
Secure Programming Cookbook for C and C++: Recipes for Cryptography, Authentication, Input Validation & More
ISBN: 0596003943
EAN: 2147483647
Year: 2006
Pages: 241

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