Why doesn't C++ have virtual variables?

Question

This might have been asked a million times before or might be incredibly stupid but why is it not implemented?

class A {       public:              A(){ a = 5;}              int a; };  class B:public A {       public:              B(){ a = 0.5;}               float a; };  int main() {     A * a = new B();      cout<<a->a;     getch();     return 0; } 

This code will access A::a. How do I access B::a?

Answer 1

To access B::a:

cout << static_cast<B*>(a)->a; 

To explicitly access both A::a and B::a:

cout << static_cast<B*>(a)->A::a; cout << static_cast<B*>(a)->B::a; 

(dynamic_cast is sometimes better than static_cast, but it can't be used here because A and B are not polymorphic.)

As to why C++ doesn't have virtual variables: Virtual functions permit polymorphism; in other words, they let a classes of two different types be treated the same by calling code, with any differences in the internal behavior of those two classes being encapsulated within the virtual functions.

Virtual member variables wouldn't really make sense; there's no behavior to encapsulate with simply accessing a variable.

Also keep in mind that C++ is statically typed. Virtual functions let you change behavior at runtime; your example code is trying to change not only behavior but data types at runtime (A::a is int, B::a is float), and C++ doesn't work that way. If you need to accommodate different data types at runtime, you need to encapsulate those differences within virtual functions that hide the differences in data types. For example (demo code only; for real code, you'd overload operator<< instead):

class A {   public:          A(){ a = 5;}          int a;          virtual void output_to(ostream& o) const { o << a; } };  class B:public A {   public:          B(){ a = 0.5;}          float a;          void output_to(ostream& o) const { o << a; } }; 

Also keep in mind that making member variables public like this can break encapsulation and is generally frowned upon.