C++ Cookbook

8.15. 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 <iostream>

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.