Question on Virtual Methods

Question

IF both methods are declared as virtual, shouldn't both instances of Method1() that are called be the derived class's Method1()?

I am seeing BASE then DERIVED called each time. I am doing some review for an interview and I want to make sure I have this straight. xD

class BaseClass
{
public:
    virtual void Method1()  { cout << "Method 1 BASE" << endl; }
};

class DerClass: public BaseClass
{
public:
    virtual void Method1() { cout << "Method 1 DERVIED" << endl; }
};


DerClass myClass;
    ((BaseClass)myClass).Method1();
    myClass.Method1();

Method 1 BASE
Method 1 DERVIED

Answer 1

No, the "C-style" cast ((BaseClass)myClass) creates a temporary BaseClass object by slicing myClass. It's dynamic type is BaseClass, it isn't a DerClass at all so the Method1 being called is the base class method.

myClass.Method1() is a direct call. As myClass is an object, not a reference there is no virtual dispatch (there would be no need).