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
In order for dynamic_cast to work the object needs to be polymorphic. The reason for this is that dynamic_cast needs somewhere to store the type information that is will use to perform the cast, and it does this by storing the information alongside the vtable for the class. In order for there to be a vtable you need to make at least one of your methods virtual.
The easiest way around this is to flag the base class destructor as virtual.
Upcasting (ie derived to base) doesn't needs a cast as the compiler is able to check that the cast would work at compile time. However, the same isn't true when downcasting.