Re: [boost] [MPL.Math] ratio

Subject: Re: [boost] [MPL.Math] ratio
From: Cromwell Enage (sponage_at_[hidden])
Date: 2008-11-14 04:06:28

--- On Thu, 11/13/08, Howard Hinnant wrote:
> On Nov 13, 2008, at 3:55 PM, Beman Dawes wrote:
> > Howard would have to comment on that; I've pinged
> > him to make him aware of this thread.
> > 
> > The Boost implementation can add extensions, and we
> > can submit a comment on the CD.  If it is as simple
> > as a typedef, it stands a good chance of being
> > accepted.  If you write it up, with a proposed
> > resolution, I'll submit it.
> 
> My only comment so far is that these look like good
> suggestions.  A defect report should show a motivating code
> sample for best chances.
> 
> -Howard

Sections: 20.4.1 [Class template ratio], 20.4.2 [Arithmetic on ratio types]

Discussion:
Let's say that client code needs to use a unary metafunction named absolute_ratio in the following type definitions, to be evaluated lazily.

    // Usage.
    typedef std::ratio<-1,2>
            negative_one_half;
    typedef absolute_ratio<negative_one_half>
            one_half;
    typedef absolute_ratio<
                std::ratio_multiplies<
                    one_half
                  , negative_one_half
                >
            >
            one_fourth;
    // End usage.

Assuming that a (partial) specialization of boost::mpl::negate exists for std::ratio, a programmer should be able to define absolute_ratio this way.

    template <typename R>
    struct absolute_ratio
      : boost::mpl::eval_if<
            R::type::num < 0
          , boost::mpl::negate<R>
          , R
        >
    {
    };

The expression R::type::num is necessary because the std::ratio_multiplies metafunction is not required to inherit from std::ratio.  However, the expression will not compile when it sees that std::ratio is not a metafunction.

Proposed resolution:
Add the nullary metafunction type definition to the ratio class template.

    namespace std {

        template <intmax_t N, intmax_t D = 1>
        class ratio
        {
         public:
            typedef ratio<N,D> type;
            static const intmax_t num;
            static const intmax_t den;
        };
    }

We should also require that each of the arithmetic metafunctions inherit from std::ratio if we want to allow the following usage (note the change from R::type::num to R::num).

    template <typename R>
    struct absolute_ratio
      : boost::mpl::eval_if<
            R::num < 0
          , boost::mpl::negate<R>
          , R
        >
    {
    };

Cheers,
Cromwell D. Enage