If a private virtual function is overridden as a public function in the derived class, what are the problems?

Question

using namespace std;
#include 
#include 

class One{
    private:
        virtual void func(){
            cout<<\"bark!\"<<         


        

Answer 1

Accessibility check is performed based on the static type of the object. The type of o is One*. This means that if One::func() is private, then o->func() won't compile.

On the other hand, which virtual member function will be called (i.e. dynamic dispatch) happens at run-time, based on the dynamic type of the object. So if One::func() is public, o->func() will call Two::func(), because o is pointing to an object of type Two.

For your sample code and use case, making One::func() private is just meaningless. But note that there's a famous idiom called Non-Virtual Interface, which makes use of private virtual member functions of base class.

---

Other suggestions:

1. Don't forget to delete o;

2. Add a virtual destructor in the base class One. Otherwise delete o; will lead to undefined behavior; e.g. the destructor of Two might not be invoked.

   class One {
       public:
           virtual ~One() {}
       // ...
   };