Question

I have some doubts about static_cast and dynamic_cast in C++. Do they completely change the object a pointer is pointing to from a class A to a class B by preserving the already-set member variables (except the ones that cannot be passed from derived to base) ?

I noticed that if I have something like

struct Base
{
    Base() { }
    virtual ~Base() { }
    virtual void Method() { cout << "Base Method"; }
};

class Derived : public Base
{
public:
    virtual void Method() { cout << "Override Method"; }
};

struct Derived2 : public Derived
{
    Derived2() { cout << "Derived2 constructor"; }
    void Method() { cout << "Override2 Method"; }
};

int main()
{       
    Base *myPointer = new Derived();    
    static_cast<Derived2*>(myPointer)->Derived2::Method();   
    delete myPointer;    
    return 0;
}

The constructor isn’t called, but the method does. How is this possible?

Answer 1

The casts don’t change the object at all. They only give you a different pointer to a related class type in the inheritance hierarchy:

Derived x;

Base         * p = &x;
AnotherClass * q = dynamic_cast<AnotherClass*>(p);

// q may or may not be NULL

For example, the above dynamic cast succeeds if we have a hierarchy AnotherClass : Base and Derived : AnotherClass (and Base is polymorphic).

A static_cast can usually be used when you already know that you have a more derived dynamic type, but happen to have only a pointer or reference to a base.

(A static cast can never be used to cast from a virtual base, in which case you always need dynamic_cast.)