Question

In the following code, while constructing obj in case 1, we construct a derived class object too, but its member functions are just inaccessible to obj. So while downcasting (i.e., in case 2), using obj as source, we have the constructed derived in it already. Why would obj need to be polymorphic?

If I confused you with my above description, why doesn’t obj need to be polymorphic when upcasting, but while downcasting it does need to be polymorphic while using dynamic_cast?

class base
{
public:
    base()
    {
        cout<< " \n base constructor \n";
    }
};

class derived : public base
{
public:
    derived()
    {
        cout << " \n derived constructor \n";
    }
};

base *obj = dynamic_cast<base*> (new derived); // case 1: explicitly upcasting
derived *OBJ = dynamic_cast<derived*> (obj);   // case 2: error

Answer 1

From 5.2.7/1 [expr.dynamic.cast] :

The result of the expression dynamic_cast<T>(v) is the result of converting the expression v to type
T.

[…]

If T is “pointer to cv1 B” and v has type “pointer to cv2 D” such that B is a base class of D, the result is a
pointer to the unique B sub-object of the D object pointed to by v.

[…]

Otherwise, v shall be a pointer to or an lvalue of a polymorphic type.

The standard even provides the following example which illustrates that the polymorphic type requirement does not stand for derived to base conversion :

struct B {};
struct D : B {};
void foo(D* dp)
{
    B* bp = dynamic_cast<B*>(dp); // equivalent to B* bp = dp;
}