Calling a Superclass Virtual FunctionProblemYou 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. SolutionQualify 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
}
DiscussionMaking 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. |