[boost] Re: Partially specializing less<> with nested template class

From: Jonathan Turkanis (technews_at_[hidden])
Date: 2004-07-08 16:59:35

"David B. Held" <dheld_at_[hidden]> wrote in message
news:cckeh3$fd0$1_at_sea.gmane.org...
> Jonathan Turkanis wrote:
> > [...]
> > Could you pull the definition of nested outside of sp
> >
> >     template<typename P1, typename P2, typename T>
> >     struct sp_nested { };
> >
> > then use a metafunction within sp:
> >
> >     template<typename P1, typename P2>
> >     struct sp {
> >          template<typename T>
> >          struct nested {
> >              typedef sp_nested<P1, P2, T> type;
> >          };
> >     };
> > [...]
>
> For syntax reasons, this is an undesirable solution.

I 'm not sure I see the syntax problem; if you mean that it's more
cumbersome to use a metafunction, I agree.

> Although,
> you could get pretty close by modifying it a little:
>
>       template<typename P1, typename P2>
>       struct sp {
>            template<typename T>
>            struct nested : sp_nested<P1, P2, T>
>            { };
>       };

Then specializing less< ... > for sp_nested wouldn't affect the
implementation of less<nested>.

<snip>

> The problem is that sp_nested<> would be an implementation detail
> that I would not want to expose to the user.  So I would not want
> the user to have to mention it in order to instantiate less<>.

I don't see the distinction between

   template <class P1, class P2, typename T>
   struct less<sp<P1, P2>::nested<T> >
        : binary_function<...>
   {
     bool operator()(...);
   };

and

   template <class P1, class P2, typename T>
   struct less<sp_nested<P1, P2, T> >
        : binary_function<...>
   {
     bool operator()(...);
   };

with respect to exposing implementation details.

Jonathan