Re: [PATCH v4] libstdc++: Remove unnecessary "& 1" from year_month_day_last::day().

[Jonathan Wakely <jwakely@redhat.com>]

On Sat, 11 Nov 2023 at 16:46, Cassio Neri <cassio.neri@gmail.com> wrote:
>
> When year_month_day_last::day() was implemented, Dr. Matthias Kretz realised
> that the operation "& 1" wasn't necessary but we did not patch it at that
> time. This patch removes the unnecessary operation.
>
> libstdc++-v3/ChangeLog:
>
>         * include/std/chrono: Remove &1 from year_month_day_last::day().
> ---
> Previous versions of this patch failed to apply. I hope it works this time.
>
> OK for trunk?

Yes, thanks - pushed to trunk.


>
>  libstdc++-v3/include/std/chrono | 24 ++++++++++++++----------
>  1 file changed, 14 insertions(+), 10 deletions(-)
>
> diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
> index 10e868e5a03..a826982803b 100644
> --- a/libstdc++-v3/include/std/chrono
> +++ b/libstdc++-v3/include/std/chrono
> @@ -1800,22 +1800,26 @@ _GLIBCXX_BEGIN_NAMESPACE_VERSION
>        {
>         const auto __m = static_cast<unsigned>(month());
>
> -       // Excluding February, the last day of month __m is either 30 or 31 or,
> -       // in another words, it is 30 + b = 30 | b, where b is in {0, 1}.
> +       // The result is unspecified if __m < 1 or __m > 12.  Hence, assume
> +       // 1 <= __m <= 12.  For __m != 2, day() == 30 or day() == 31 or, in
> +       // other words, day () == 30 | b, where b is in {0, 1}.
>
> -       // If __m in {1, 3, 4, 5, 6, 7}, then b is 1 if, and only if __m is odd.
> -       // Hence, b = __m & 1 = (__m ^ 0) & 1.
> +       // If __m in {1, 3, 4, 5, 6, 7}, then b is 1 if, and only if, __m is
> +       // odd.  Hence, b = __m & 1 = (__m ^ 0) & 1.
>
> -       // If __m in {8, 9, 10, 11, 12}, then b is 1 if, and only if __m is even.
> -       // Hence, b = (__m ^ 1) & 1.
> +       // If __m in {8, 9, 10, 11, 12}, then b is 1 if, and only if, __m is
> +       // even.  Hence, b = (__m ^ 1) & 1.
>
>         // Therefore, b = (__m ^ c) & 1, where c = 0, if __m < 8, or c = 1 if
>         // __m >= 8, that is, c = __m >> 3.
>
> -       // The above mathematically justifies this implementation whose
> -       // performance does not depend on look-up tables being on the L1 cache.
> -       return chrono::day{__m != 2 ? ((__m ^ (__m >> 3)) & 1) | 30
> -                                   : _M_y.is_leap() ? 29 : 28};
> +       // Since 30 = (11110)_2 and __m <= 31 = (11111)_2, the "& 1" in b's
> +       // calculation is unnecessary.
> +
> +       // The performance of this implementation does not depend on look-up
> +       // tables being on the L1 cache.
> +       return chrono::day{__m != 2 ? (__m ^ (__m >> 3)) | 30
> +         : _M_y.is_leap() ? 29 : 28};
>        }
>
>        constexpr
> --
> 2.41.0
>