Virtual methods allow the subclass methods to be called even if the pointer is of type base class. The code snippet is taken from Stack Overflow:
class Base
{
public:
void Method1() {
std::cout << "Base::Method1" << std::endl;
}
virtual void Method2() {
std::cout << "Base::Method2" << std::endl;
}
};
class Derived : public Base
{
public:
void Method1() {
std::cout << "Derived::Method1" << std::endl;
}
void Method2() {
std::cout << "Derived::Method2" << std::endl;
}
};
// Note - constructed as Derived, but pointer stored as Base*
Base* obj = new Derived ();
obj->Method1 (); // Prints "Base::Method1"
obj->Method2 (); // Prints "Derived::Method2"If the virtual qualifier is missing from Method2 then we would have early binding and the only the method in the base class would be available.
class Base
{
public:
void Method1() {
std::cout << "Base::Method1" << std::endl;
}
void Method2() {
std::cout << "Base::Method2" << std::endl;
}
};
class Derived : public Base
{
public:
void Method1() {
std::cout << "Derived::Method1" << std::endl;
}
void Method2() {
std::cout << "Derived::Method2" << std::endl;
}
}
// Note - constructed as Derived, but pointer stored as Base*
Base* obj = new Derived();
obj->Method1(); // Prints "Base::Method1"
obj->Method2(); // Prints "Base::Method2"