Virtual functions (C++ only)

The following examples demonstrate the differences between static and dynamic binding. The first example demonstrates static binding:

#include <iostream>
using namespace std;

struct A {
   void f() { cout << "Class A" << endl; }
};

struct B: A {
   void f() { cout << "Class B" << endl; }
};

void g(A& arg) {
   arg.f();
}

int main() {
   B x;
   g(x);
}

The following is the output of the above example:

Class A

When function g() is called, function A::f() is called, although the argument refers to an object of type B. At compile time, the compiler knows only that the argument of function g() will be a reference to an object derived from A; it cannot determine whether the argument will be a reference to an object of type A or type B. However, this can be determined at runtime. The following example is the same as the previous example, except that A::f() is declared with the virtual keyword:

#include <iostream>
using namespace std;

struct A {
   virtual void f() { cout << "Class A" << endl; }
};

struct B: A {
   void f() { cout << "Class B" << endl; }
};

void g(A& arg) {
   arg.f();
}

int main() {
   B x;
   g(x);
}

The following is the output of the above example:

Class B

The virtual keyword indicates to the compiler that it should choose the appropriate definition of f() not by the type of reference, but by the type of object that the reference refers to.

You redefine a virtual member function, like any member function, in any derived class. Suppose you declare a virtual function named f in a class A, and you derive directly or indirectly from A a class named B. If you declare a function named f in class B with the same name and same parameter list as A::f, then B::f is also virtual (regardless whether or not you declare B::f with the virtual keyword) and it overrides A::f. However, if the parameter lists of A::f and B::f are different, A::f and B::f are considered different, B::f does not override A::f, and B::f is not virtual (unless you have declared it with the virtual keyword). Instead B::f hides A::f. The following example demonstrates this:

#include <iostream>
using namespace std;

struct A {
   virtual void f() { cout << "Class A" << endl; }
};

struct B: A {
   void f(int) { cout << "Class B" << endl; }
};

struct C: B {
   void f() { cout << "Class C" << endl; }
};

int main() {
   B b; C c;
   A* pa1 = &b;
   A* pa2 = &c;
//   b.f();
   pa1->f();
   pa2->f();
}

The following is the output of the above example:

Class A
Class C

The function B::f is not virtual. It hides A::f. Thus the compiler will not allow the function call b.f(). The function C::f is virtual; it overrides A::f even though A::f is not visible in C.

The return type of an overriding virtual function may differ from the return type of the overridden virtual function. This overriding function would then be called a covariant virtual function. Suppose that B::f overrides the virtual function A::f. The return types of A::f and B::f may differ if all the following conditions are met:

• The function B::f returns a reference or pointer to a class of type T, and A::f returns a pointer or a reference to an unambiguous direct or indirect base class of T.
• The const or volatile qualification of the pointer or reference returned by B::f has the same or less const or volatile qualification of the pointer or reference returned by A::f.
• The return type of B::f must be complete at the point of declaration of B::f, or it can be of type B.

The following example demonstrates this:

#include <iostream>
using namespace std;

struct A { };

class B : private A {
   friend class D;
   friend class F;
};

A global_A;
B global_B;

struct C {
   virtual A* f() {
      cout << "A* C::f()" << endl;
      return &global_A;
   }
};

struct D : C {
   B* f() {
      cout << "B* D::f()" << endl;
      return &global_B;
   }
};

struct E;

struct F : C {

//   Error:
//   E is incomplete
//   E* f();
};

struct G : C {

//   Error:
//   A is an inaccessible base class of B
//   B* f();
};

int main() {
   D d;
   C* cp = &d;
   D* dp = &d;

   A* ap = cp->f();
   B* bp = dp->f();
};

The following is the output of the above example:

B* D::f()
B* D::f()

The statement A* ap = cp->f() calls D::f() and converts the pointer returned to type A*. The statement B* bp = dp->f() calls D::f() as well but does not convert the pointer returned; the type returned is B*. The compiler would not allow the declaration of the virtual function F::f() because E is not a complete class. The compiler would not allow the declaration of the virtual function G::f() because class A is not an accessible base class of B (unlike friend classes D and F, the definition of B does not give access to its members for class G).

A virtual function must be one of the following:

• Defined
• Declared pure
• Defined and declared pure

A base class containing one or more pure virtual member functions is called an abstract class.

Related information

• Abstract classes (C++ only)