How does dynamic casting works when you don't cast to the most derived

Question

Lets examine following case

struct A{ virtual ~A(){} }; struct B : public A{ virtual ~B(){} }; struct C : public B{ virtual ~C(){} }; int main(){ A* a = new C(); B* b = dynamic_cast<B*>(a); }

How does dynamic_cast know that B is a super class of C at the runtime. I understand that dynamic_cast accesses type_info of *a and finds that *a is actually of type C by examining name property. But how not having all the information that compilator has about classes inheritance does dynamic_cast know that B is a superclass of C having only information that the type of *a is C? Does that make any sense?

What makes you think that the RTTI doesn't have all the type hierarchy information? — Alan Birtles
– Alan Birtles, Commented Nov 21, 2020 at 18:23

Kuba hasn't forgotten Monica · Accepted Answer · 2020-11-21 18:30:52Z

The premise is wrong: dynamic_cast does indeed have all the information it needs. The compiler encodes the type information in read-only type information records that are a part of the binary while it executes. The vtable pointer embedded in each polymorphic object allows dynamic_cast to find this “baked in” type information, since the virtual method table contains more information than just method pointers. That’s the common way it’s implemented, and also a matter of a fixed ABI on x86-64, ie all compilers for that architecture encode this information the same iirc.

But vtable holds only type_info object which doesn't have anything but exact type in it. Or is it a key to some entry in class map that holds list of whole hierarchy that is not accessible for casual user?

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