Question

Lets say we have the following two class definitions.

#include <iostream>
#include <array>

class A
{
public:
  virtual void f() = 0;
};

class B : public A
{
public:
  virtual void f() { std::cout << i << std::endl; }
  int i;
};

Here sizeof(B) == 8, presumably 4 the virtual pointer and 4 for the int.

Now lets say we make an array of B, like so:

std::array<B, 10> x;

Now we get sizeof(x) == 80.

If my understanding is correct, all method calls on elements of x are resolved statically, as we know the type at compile time. Unless we do something like A* p = &x[i] I don’t see a need to even store the virtual pointer.

Is there a way to create an object of type B without a virtual pointer if you know it is not going to be used?

i.e. a template type nonvirtual<T> which does not contain a virtual pointer, and cannot be pointed to by a subtype of T?

Answer 1

Is there a way to create an object of type B without a virtual pointer if you know it is not going to be used?

No. Objects are what they are. A virtual object is virtual, always.

After all, you could do this:

A *a = &x[2];
a->f();

That is perfectly legitimate and legal code. And C++ has to allow it. The type B is virtual, and it has a certain size. You can’t make a type be a different type based on where it is used.