Re: [committed] libstdc++: Suppress pedantic warnings about __int128

[Jonathan Wakely <jwakely@redhat.com>]

On Fri, 16 Jul 2021 at 09:40, Jonathan Wakely <jwakely.gcc@gmail.com> wrote:
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> On Fri, 16 Jul 2021, 09:38 Jonathan Wakely, <jwakely.gcc@gmail.com> wrote:
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>> On Fri, 16 Jul 2021, 09:30 Jakub Jelinek via Libstdc++, <libstdc++@gcc.gnu.org> wrote:
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>>> On Fri, Jul 16, 2021 at 10:27:09AM +0200, Jakub Jelinek via Gcc-patches wrote:
>>> > On Fri, Jul 16, 2021 at 08:41:06AM +0100, Jonathan Wakely via Gcc-patches wrote:
>>> > > --- a/libstdc++-v3/include/bits/max_size_type.h
>>> > > +++ b/libstdc++-v3/include/bits/max_size_type.h
>>> > > @@ -417,7 +417,10 @@ namespace ranges
>>> > >  #endif
>>> > >
>>> > >  #if __SIZEOF_INT128__
>>> > > +#pragma GCC diagnostic push
>>> > > +#pragma GCC diagnostic ignored "-Wpedantic"
>>> > >        using __rep = unsigned __int128;
>>> > > +#pragma GCC diagnostic pop
>>> >
>>> > At least in simple cases like this, wouldn't
>>> >       using __rep = __extension__ unsigned __int128;
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>>> __extension__ using __rep = unsigned __int128;
>>> actually (now tested).
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>> Ah, thanks. I didn't find the right syntax, and I know __extension__ doesn't work in other cases, like quad float literals, so I assumed it doesn't work here. I suppose the literals don't work because the warning comes from the processor, which doesn't understand __extension__ (and also ignores the diagnostic pragma).
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> That grammar for a using-declaration makes no sense at all btw ;-)

Hmm, in fact it seems that we can just use the __uint128_t typedef
instead, which doesn't give a pedwarn:

  using __rep = __uint128_t;

Is that typedef always available if __int128 is? There's a comment in
gcc/c-family/c-common.c that I don't understand:

#if HOST_BITS_PER_WIDE_INT >= 64
  /* Note that this is different than the __int128 type that's part of
     the generic __intN support.  */
  if (targetm.scalar_mode_supported_p (TImode))
    lang_hooks.decls.pushdecl (build_decl (UNKNOWN_LOCATION,
                       TYPE_DECL,
                       get_identifier ("__int128_t"),
                       intTI_type_node));
#endif

They are the same type in C++, so what is "different"? Is it possible
for __int128 to be different from a TImode integer?

We can still use __extension__ elsewhere, for defining explicit
specializations using the non-standard integers, e.g.

#define __INT_N(TYPE)                  \
+  __extension__                                \
  template<>                           \
    struct __is_integer<TYPE>          \
    {                                  \
      enum { __value = 1 };            \
      typedef __true_type __type;      \
    };