Question

I have a problem in virtual function:
Here is some code as an example:

class A
   {
      public : virtual  void print(void)
           {
              cout<< "A::print()"<<endl;
           }
    };
 class B : public A
    {
      public : virtual void print(void)
           {
               cout<<"B::print()"<<endl;
           }
    };
 class C : public A
    {
      public : void print(void)
            {
               cout<<"C::print()"<<endl;
            }
     };
  int main(void)
     {
         A a,*pa,*pb,*pc;
         B b;
         C c;
         pa=&a;
         pb=&b;
         pc=&c;

         pa->print();
         pb->print();
         pc->print();

         a=b;
         a.print();
         return 0;
       }

the result:
A::print()
B::print()
C::print()
A::print()

I know it’s a Polymorphism ,and know have a table called virtual-function-table,but I don’t know how it is to achieve,And

   a=b;
   a.print();

the result is: A::print() not B::print(),why it hasn’t polymorphism.
thank you!

Answer 1

a=b;
a.print();

It will print A::print() because a=b causes object-slicing, which means a gets only the a-subobject of b. Read this :

• What is object slicing? .

Note that runtime-polymorphism can be achieved only through pointer and reference types. In the above code, a is neither pointer, nor reference type:

A * ptr = &b; //syntax : * on LHS, & on RHS
A & ref =  b; //syntax : & on LHS, that is it!

ptr->print(); //will call B::print() (which you've already seen)
ref.print();  //will call B::print() (which you've not seen yet)