How does C++ compiler know the size of a object with superclasses

Question

Exactly as the question states!

Let's say I have the following snippet

class A
{
    int x;
    int y;
}

class B : public A
{
    int z;
}

class C
{
    A a;

    public C(A a) : a(a){}
}

What would happen if I called C's constructor with a B class, would it copy it's A part data in the class? Or also keep it's B data somewhere?

Thanks in advance! It might be a stupid question but I never understood actually.

Answer 1

If you pass an instance of B to the C constructor that takes an A by value, the B instance will be sliced, and just the A part will remain. So :

would it copy it's A part data in the class?

this.

So, specifically, there is no way to turn the C::a member back into a B instance with the same value for z as the original B instance - that information has been lost during the (irreversible) slicing operation.

Answer 2

What would happen if I called C's constructor with a B class, would it copy it's A part data in the class?

Yes. This is known as slicing - the argument is created using A's copy constructor, which just copies the A subobject.

Or also keep it's B data somewhere?

No.