Re: [boost] [range] How to extend Boost.Range?

From: Rob Stewart (stewart_at_[hidden])
Date: 2005-09-20 16:29:51

From: David Abrahams <dave_at_[hidden]>
> Rob Stewart <stewart_at_[hidden]> writes:

> > I've been thinking about this issue and wonder if the
> > following wouldn't solve the problem neatly for both
> > conforming and non-conforming compilers:
> >
> > // whatever/foo.h
> > namespace whatever
> > {
> >    class foo
> >    {
> >       ...
> >    };
> >    // for conforming compilers
> >    unspecified begin(foo const &)
> >    {
> >       ...
> >    };
> >    // for non-conforming compilers
> >    namespace boost { namespace range {
> >       using whatever::begin;
> >    } }
> >    // boost/range/begin.hpp
> >    namespace boost
> >    {
> >       inline template <T>
> >       unspecified begin(T const & range_i)
> >       {
> > 	 using boost::range::begin;
> > 	 return begin(range_i);
> >       }
> >    }

> Careful.  Many of the problems you can run into with
> dispatching onlyshow up in the case where the definition of
> begin you want to beselected *follows* the template that uses
> it.

Right.  That was the case you'd been discussing and I missed its
application here.  However, if the author of whatever::foo
defines, as a matter of convention or coercion,* begin() in
namespace boost::range in the same header as foo, and doesn't
invoke it before it's definition, then the problem order won't
occur.

> > Did I miss something?

> Two-phase name lookup. 
>   14.6.4 Dependent name resolution [temp.dep.res]

LOL It seems you like quoting that now that you've found it!  I
do know about two-phase lookup, but I don't have to deal with it
often so I don't always recognize the trouble it can cause a
solution like this.

Nevertheless, won't the problem only occur for non-conforming
compilers and only when the definition of
boost::range::begin(foot const &) is in a separate header
(provided the author/maintainer of foo follows the prescribed
course when it's in the same header)?

> Unless you arrange for boost::range to be an associated namespace offoo, you're setting up dangerous #include order dependencies.

Of course.

* It would be documented as the only correct thing to do were
  this an accepted workaround for non-conforming compilers.

-- 
Rob Stewart                           stewart_at_[hidden]
Software Engineer                     http://www.sig.com
Susquehanna International Group, LLP  using std::disclaimer;