From: Joel de Guzman (joel_at_[hidden])
Date: 2008-08-22 01:04:43
Zeljko Vrba wrote:
> I have succeeded in writing the following transform functor:
>
> typedef bf::map<bf::pair<net, unsigned>, bf::pair<pin, unsigned> > Map;
> typedef bf::vector<unsigned, unsigned> Pair;
>
> struct Map2Pair
> {
> template<typename Sig>
> struct result;
>
> template<typename U>
> struct result<Map2Pair(U)>
> : remove_reference<T::second_type>
> { };
>
> template<typename T>
> typename T::second_type operator()(T t) const
> {
> return t.second;
> }
> };
>
> I have tested it, it works. The result magic has been copied from the
> transform_view example in the documentation and adjusted for the proper return
> type. Where can I read a bit more about result conventions?
Read on Boost.Result of and the TR1 result_of conventions.
http://www.boost.org/doc/libs/1_36_0/libs/utility/utility.htm
http://anubis.dkuug.dk/jtc1/sc22/wg21/docs/papers/2003/n1454.html
In particular,
> how does the
>
> template<typename U>
> struct result<Map2Pair(U)>
> : remove_reference<T::second_type>
> { };
>
> specialization work when Map2Pair is a struct, and it doesn't even have a
> constructor!
It's not a function call. It's a function declarator. This is first
discussed and put into use in the Boost.Function library:
http://www.boost.org/doc/libs/1_36_0/doc/html/function/tutorial.html#id2903300
> Even worse, why does it work at all when T is not in the
> scope of the template (.. because it's never used?)
Which template?
Yes, it is not used,
> the following functor also works:
>
> struct Map2Pair
> {
> template<typename T>
> typename T::second_type operator()(T t) const
> {
> return t.second;
> }
> };
>
> When and how is the above version preferred than the simpler version?
Are you sure the one above works? You need the result type,
AFAICT.
Regards,
-- Joel de Guzman http://www.boostpro.com http://spirit.sf.net