c++ inheritance, vtable after casting

given the next simple example,

#include <iostream>
using namespace std;

class A {
    public :
    void print() {
        cout<<"A print"<<endl;
    }
};

class B : public A {
    void virtual print() {
        cout<<"B print"<<endl;
    }

};

int main() {
    A* aPointerB=new B();
    aPointerB->print();
    return 0;
}

the output is : A Print

from my understanding :

new B()

creates an object on the heap with a virtual table that contains one method and that's B::print(); i h开发者_开发问答ave two ideas to why this is happening,

1. when casting the pointer to A is the virtual table entry for B::print() removed.
2. in runtime the program doesn't check the vtable entry for print() because we are of type A and print() is not virtual in A, therefor it runs A::print().

which one is it ?

EDIT

it seems that my question wasn't clear, i know that A::print() is running because print() isn't virtual in A, my question is which of the 2 options is it for the reason print is running in A understanding the considering the way vtables and casting work.

thanks !

It happens because you didn't define print as virtual in A. That means that when calling on A objects, the lookup is not virtual.

A's vtable doesn't contain any print function, since it's not virtual. For this reason if you call print on an object of type A the vtable isn't checked, and A::print (which is not virtual) gets called.

This means the code generated by the compiler just calls A::print, without even accessing the vtable.

Between your two ideas the second is the right one; the first one makes no sense: vtables are never modified at runtime.

The print() in A needs to be declared as virtual, not the one in B as you have it.