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From: "redi at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug libstdc++/110719] New: Should chrono formatters always use
 std::time_put for locale's representation?
Date: Tue, 18 Jul 2023 14:43:45 +0000
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X-Bugzilla-Product: gcc
X-Bugzilla-Component: libstdc++
X-Bugzilla-Version: 13.1.1
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D110719

            Bug ID: 110719
           Summary: Should chrono formatters always use std::time_put for
                    locale's representation?
           Product: gcc
           Version: 13.1.1
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: libstdc++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: redi at gcc dot gnu.org
  Target Milestone: ---

auto t =3D std::chrono::system_clock::now();
  auto loc =3D std::locale::classic();
  std::cout << std::format(loc, "{:%EX %OS}\n", t);

This prints something like:

14:30:46.809059031 46

The %EX output is produced by calling std::format again with a format string
based on the locale's D_FMT, which for the C locale is something like:
%H:%M:%S. And using std::format("{:%H:%M:%S}", t) prints fractional seconds=
 for
the %S part.

The %OS output is produced by calling std::time_put::put with the %OS format
string and a struct tm with tm_sec set to the integer number of seconds. Th=
is
doesn't print the fractional part.

If chrono::parse("%EX", t) uses std::time_get then this presents a problem =
for
round-tripping, as the formatted output will have fractional seconds, but t=
he
parsed input will not consume that fractional part.

Should we consistently use std::time_put for all locale-specific output?
Alternatively, we could use time_point_cast and duration_cast to round to
seconds. None of the locale-specific formats print fractional seconds.

It would be useful to profile std::format with and without std::time_get, to
see if reusing std::format performs better. If it doesn't, using std::time_=
put
might be simpler.=