Re: How dynamic_cast works internally?

On Oct 22, 4:51 am, Bo Yang <struggl...@gmail.com> wrote:


Consider this case:
class A{virtual ~A(){};};
class B:public A{};
A*a=new B;
B*b =dynamic_cast<B*>(a);
Conceptually, it is like this if is were written in C:

void A_dtor(A* ){}
typeinfo A_RTTI(){
    static typeinfo
   return typeinfo("A");
}//compiler generated
void(*)() A_vtbl[]={A_dtor,A_RTTI};
struct A{
      A_vtbl*vptr;
};
A_ctor(){vptr=A_vtbl;}//compiler generated

void B_dtor(B* ){}
typeinfo B_RTTI(){
    static typeinfo
   return typeinfo("B");
}//compiler generated
void(*)(B*) B_vtbl[]={B_dtor,B_RTTI};
struct B{
      B_vtbl*vptr;
};
B_ctor(){vptr=B_vtbl;}//compiler generated

This is what A and B conceptually look like underneath the hood. Now
when dynamic_cast is called, the compiler generates a function
something like

B* Dynamic_Cast(A*a ){
    if(*((a->vptr)+1)()==B_RTTI())
       return a;
    return 0;
}


for references.

You can see the reason now why dynamic_cast only works for types that
have "virtual" somewhere, either a function or inheritance.

Lance