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From: Joaquín Mª López Muñoz (joaquin_at_[hidden])
Date: 2005-09-29 08:18:35

Daryle Walker ha escrito:

> On 9/26/05 10:49 AM, "Joaquín Mª López Muñoz" <joaquin_at_[hidden]> wrote:
> > Martin Bonner ha escrito:
> >
> >> ----Original Message----
> >> From: Joaquin M Lopez Munoz [mailto:joaquin_at_[hidden]]
> >>
> >>> Is BOOST_HAS_INTPTR_TYPES supposed to inform whether there are
> >>> vendor-provided ::intrptr_t/::uintrptr_t types? If so, I fail to see its
> >>> utility, since boost::intrptr_t/boost::untrptr_t can be provided
> >>> universally.
> >>
> >> Not if the compiler doesn't HAVE a integral type which can hold a pointer.
> >> For example, a compiler for Intel processors with 32 bit longs, no long long,
> >> and 16:32 pointers.
> >
> > But then, one doesn't really care whether the vendor has ::[u]intrptr_t as
> > long as she has Boost-provided boost::[u]intrptr_t, right? To cope with the
> > situtations you describe, the macro should be named then something like
> > that makes it clear we are refererring to boost/cstdint.hpp rather than the
> > vendor stuff.
> You assuming that Martin's scenario is caused by the compiler maker being
> unwilling to map a pointer type to an integral type. The actual point was
> that the compiler maker may be UNABLE to create a mapping. In other words,
> we (i.e. Boost) can't make the typedef-s either! (Unless you're so slick
> you can make a mapping even if "sizeof( void * ) > sizeof( [u]intmax_t )".)

I think I failed to express myself. In fact, I agree with you. Let's see,
a vendor may not provide ::[u]intrptr_t for the following reasons:

a) they didn't bother to do it (as this is not mandated by the C++ std anway)
b) it can't be done, because there's no such integral type.

So, what I propose is that boost::[u]intrptr_t is provided whenever
possible (regardless of the vendor's headers) and that the macro
situation b).

AFAICS this also reflects your position, doesn't it?

Best regards,

Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo

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