Re: size of class same as base class if no new data members?

From:
Victor Bazarov <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Thu, 09 Apr 2009 10:46:16 -0400
Message-ID:
<grl1np$3qc$1@news.datemas.de>
guy.tristram@gmail.com wrote:

Can I assume that if I have:

class A
{
   /* a few built in types */


You mean, there are data members here? 'cause your class looks like it
has no data members at all...

};

class B : public A
{
private:
   //prevent copy and assignment
   B( B const & );
   operator=( B const & );
};

instances of class A will be the same size as instances of class B?


Since 'B' adds no data and no virtual functions or virtual base classes,
then yes, most likely.

More particularly, if I have:

   std::vector< A > v(2);
   B *b = static_cast< B * >( &v[0] );
   ++b;

will b point to the second element of v?


No. The code has undefined behaviour. You can't use 'static_cast' that
way. The object at the address &v[0] is of type 'A', and it is not a
subobject of some other object of type 'B'.

 > Does the standard offer any

guarantees?


Of course not.

 > My motivation is that I want to be able to grow a vector

of As, but prevent clients from copying them thereafter.


It would seem that you're solving a wrong problem. If you care to
explain what you're trying to accomplish, folks here might have
suggestions on what you should do to get it.

 > I realize

that I am then using As as though they were Bs and I should perhaps
inherit the other way around, making the copy constructor and
assignment operator protected in the client version, then re-
publicizing them in the internal version, but that would mean
explicitly implementing them.


Uh... Don't know.

V
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