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.

[The Definitive C++ Book Guide and List](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). No excuse. — Ivan Aksamentov - Drop, Dec 06 '13 at 13:34

Sander De Dycker · Accepted Answer · 2013-12-06T13:57:01.890

5

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.

edited Dec 06 '13 at 13:57

answered Dec 06 '13 at 13:35

Sander De Dycker

16,053
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Thanks a lot! It helps me becoming a master in C++:P – Dec 06 '13 at 13:38

score 3 · Answer 2 · answered Dec 06 '13 at 13:35

3

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.

answered Dec 06 '13 at 13:35

Mike Seymour

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Thanks a lot! Still need to wait 8 minutes to accept an answer though:P – Dec 06 '13 at 13:38

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

2 Answers2