Re: Boost and multi-level inheritance

From:
David Abrahams <dave@boost-consulting.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 28 Aug 2007 16:03:00 CST
Message-ID:
<87absb4j9k.fsf@grogan.peloton>
on Tue Aug 28 2007, Marcin Gil <marcin.gil-AT-NOSPAMgmail.com> wrote:

Hi!

Does it produce much overhead if i.e. I had:

class A: public boost::less_than_comparable<A>;

class B: public A, public boost::less_than_comparable<B>;

class C: public B, public boost::less_than_comparable<C>;

?


What kind of overhead do you anticipate? Some compilers will allocate
a little extra space in B and C to store the empty multiply-inherited
less_than_comparable<B/C> instances.

you can get around that using

  class A: public boost::less_than_comparable<A>
  {
     bool operator<(A const&) const;
  };

  class B: public boost::less_than_comparable<B,A>;
  {
     bool operator<(B const&) const;
  };

  class C: public boost::less_than_comparable<C,B>
  {
     bool operator<(C const&) const;
  };

see http://boost.org/libs/utility/operators.htm#chaining

Let's say that the type A is not directly used and is only
used as a pointer/reference parameter type in other classes, ie.:

class D
{
   public:
     int fun(const A&);
}

and having:
B b;
C c;
D d;

should work for
d.fun(b);
d.fun(c);


OK...

Also I need this operation:
b < c
to work.


It will "work" by using B::operator<(B const&). Whether or not that's
the function you want it to call, I cannot say.

Moreover, for:
B b1; A& ba1 = b1;
B b2; A& ba2 = b2;

the
ba1 < ba2
should also work with correct operator<().


Which is the correct one? You haven't explained the semantics you
want. If you expect that to call B's operator<, then you'd better
make operator< virtual in A.

--
Dave Abrahams
Boost Consulting
http://www.boost-consulting.com

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