Question

I know about the basic concept of virtual function and run-time call. But i tried
running some piece of code which confused me

   class A {
   public:
    A& operator=(char) {
      cout << "A& A::operator=(char)" << endl;
      return *this;
    }
    virtual A& operator=(const A&) {
      cout << "A& A::operator=(const A&)" << endl;
      return *this;
    }
   };

   class B : public A {
   public:
      B& operator=(char) {
        cout << "B& B::operator=(char)" << endl;
        return *this;
      }

      virtual B& operator=(const B&) {
        cout << "B& B::operator=(const B&)" << endl;
        return *this;
      }
   };

   int main() {
    B b1;
    B b2;
    A* ap1 = &b1;
    A* ap2 = &b1;
    *ap1 = 'z';
    *ap2 = b2; 
   }

Running this program give me the following output:-

   A& A::operator=(char)  //expected output
   A& A::operator=(const A&) //Why this Output?  in case of  *ap2 = b2;

b2 is an object of B type but still it goes in virtual A& operator=(const A&)
and not virtual B& operator=(const B&). Why is this so ?

Answer 1

Because virtual B& operator=(const B&) does not override virtual A& operator=(const A&); the arguments are different.