Re: [boost] Re: should boost::ref forward some operators?

From: Victor A. Wagner, Jr. (vawjr_at_[hidden])
Date: 2004-02-03 18:18:41

At Tuesday 2004-02-03 14:47, you wrote:
>Neal D. Becker wrote:
> > I ran into the following problem.  I have an algorithm which looks
> > like:
> >
> > template<typename in_t, typename out_t>
> > void Alg (in_t in, in_t inend, out_t out) {
> >   for (; in != inend; in++, out++)
> >     do_something()
> > }
> >
> > Now if I want to call this algorithm more than once, it would be
> > convenient to pass out by ref, so it would be incremented:
> >
> > Alg (v.begin(), v.end(), boost::ref (out))
> > that is, pass a reference to some kind of output iterator
> >
> > this fails, because boost::ref doesn't have operator++ defined for
> > reference_wrapper.
> >
> > My question is, should boost::ref forward operators like ++ so that
> > this would work?
> >
>Altough I kindof see your point, I still feel your key problem is that your
>algorithm "should" return the used iterator? (Ala functional/STL style).
>
>template<typename in_t, typename out_t>
>out_t Alg (in_t in, in_t inend, out_t out) {
>    for (; in != inend; in++, out++)
>       do_something();
>    return out;
>}

perhaps also    ++in, ++out
since they're never slower and often faster

btw, what's the difference between this and std::transform()? (unless it's 
the constant do_something();

>Regards
>// Fredrik Blomqvist
>
>
>
>_______________________________________________
>Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost

Victor A. Wagner Jr.      http://rudbek.com
The five most dangerous words in the English language:
               "There oughta be a law"