c++ - Is the whole object freed with a non-virtual destructor and a Base class pointer?

Question

If a Base class does not have a virtual destructor (in order to avoid the vtable entry for instance) and the Derived class has only basic attributes, does it free all the memory allocated by new, when the pointer of the Base class is deleted? I know the destructor of the Derived class will not be called but I am wondering if the memory allocated by the whole object will be freed? I assume also that calling delete on a Derived pointer will free the whole memory space.

Also, if it does not free the Derived class part of the memory, how does it work in the same case but with of a virtual destructor in the Base class, to know how much memory to free?

Exemple:

class Base {
  public:
    int a;
    int b;
   Base() {}
  ~Base() {}
};

class Derived : public Base {
  public:
    int c;
    int d;
    Derived() {}
    ~Derived() {}
};

int main() {
  Base *p = new Derived();
  delete p; // is memory allocated for Derived freed?
  return 0;
}

Answer 1

It's undefined behavior, so anything can happen. Quote from the standard [expr.delete]:

In the first alternative (delete object), if the static type of the object to be deleted is different from its dynamic type, the static type shall be a base class of the dynamic type of the object to be deleted and the static type shall have a virtual destructor or the behavior is undefined.

Although how it works is the detail of implementation, the typical implementation may override automatically the destructor in the derived classes and implement the release of the memory there. Note that you must define the virtual destructor in the base class so the implementation can reserve an entry in the virtual table.