By default, C++ matches a function call with the correct function definition at compile time. This is called static binding. You can specify that the compiler match a function call with the correct function definition at run time; this is called dynamic binding. You declare a function with the keyword virtual if you want the compiler to use dynamic binding for that specific function.
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 AWhen 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 run time. 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 BThe 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.
Therefore, a virtual function is a member function you may redefine for other derived classes, and can ensure that the compiler will call the redefined virtual function for an object of the corresponding derived class, even if you call that function with a pointer or reference to a base class of the object.
A class that declares or inherits a virtual function is called a polymorphic class.
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 CThe 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.
If you declare a base class destructor as virtual, a derived class destructor will override that base class destructor, even though destructors are not inherited.
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 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 cannot be global or static because, by definition, a virtual function is a member function of a base class and relies on a specific object to determine which implementation of the function is called. You can declare a virtual function to be a friend of another class.
If a function is declared virtual in its base class, you can still access it directly using the scope resolution (::) operator. In this case, the virtual function call mechanism is suppressed and the function implementation defined in the base class is used. In addition, if you do not override a virtual member function in a derived class, a call to that function uses the function implementation defined in the base class.
A virtual function must be one of the following:
A base class containing one or more pure virtual member functions is called an abstract class.
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