[PATCH] libstdc++: Optimize chrono year::is_leap
Jonathan Wakely
jwakely@redhat.com
Thu Jul 30 15:11:07 GMT 2026
On Thu, 30 Jul 2026 at 01:58, Francisco Muniz wrote:
>
> Use Falk Hueffner's leap-year test for year::is_leap after shifting
> the valid std::chrono::year range by a multiple of 400. The shift
> preserves divisibility by 4, 100, and 400, and the unsigned conversion
> gives the intended modulo 2^32 arithmetic.
>
> Idea by Cassio Neri: add 32800, which is 82 * 400, to shift the signed
> year range into the supported non-negative range.
This is interesting, but your new code is slower than Cassio's code
when I benchmark it. Have you done your own benchmarking?
>
> Tested on x86_64-pc-linux-gnu:
> make -j$(nproc) all-gcc
> make -j$(nproc) all-target-libstdc++-v3
> make check RUNTESTFLAGS='conformance.exp=std/time/year/1.cc'
> make check RUNTESTFLAGS='conformance.exp=std/time/year/2.cc'
>
> libstdc++-v3/ChangeLog:
>
> * include/std/chrono (year::is_leap): Use Hueffner leap-year
> test after biasing the year by a multiple of 400.
>
> Signed-off-by: Francisco Muniz <munizfco@gmail.com>
> ---
> libstdc++-v3/include/std/chrono | 31 ++++++++-----------------------
> 1 file changed, 8 insertions(+), 23 deletions(-)
>
> diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
> index 692fd6025e7..4483914c08b 100644
> --- a/libstdc++-v3/include/std/chrono
> +++ b/libstdc++-v3/include/std/chrono
> @@ -904,29 +904,14 @@ _GLIBCXX_BEGIN_NAMESPACE_VERSION
> constexpr bool
> is_leap() const noexcept
> {
> - // Testing divisibility by 100 first gives better performance [1], i.e.,
> - // return _M_y % 100 == 0 ? _M_y % 400 == 0 : _M_y % 16 == 0;
> - // Furthermore, if _M_y % 100 == 0, then _M_y % 400 == 0 is equivalent
> - // to _M_y % 16 == 0, so we can simplify it to
> - // return _M_y % 100 == 0 ? _M_y % 16 == 0 : _M_y % 4 == 0. // #1
> - // Similarly, we can replace 100 with 25 (which is good since
> - // _M_y % 25 == 0 requires one fewer instruction than _M_y % 100 == 0
> - // [2]):
> - // return _M_y % 25 == 0 ? _M_y % 16 == 0 : _M_y % 4 == 0. // #2
> - // Indeed, first assume _M_y % 4 != 0. Then _M_y % 16 != 0 and hence,
> - // _M_y % 4 == 0 and _M_y % 16 == 0 are both false. Therefore, #2
> - // returns false as it should (regardless of _M_y % 25.) Now assume
> - // _M_y % 4 == 0. In this case, _M_y % 25 == 0 if, and only if,
> - // _M_y % 100 == 0, that is, #1 and #2 are equivalent. Finally, #2 is
> - // equivalent to
> - // return (_M_y & (_M_y % 25 == 0 ? 15 : 3)) == 0.
> -
> - // References:
> - // [1] https://github.com/cassioneri/calendar
> - // [2] https://godbolt.org/z/55G8rn77e
> - // [3] https://gcc.gnu.org/pipermail/libstdc++/2021-June/052815.html
> -
> - return (_M_y & (_M_y % 25 == 0 ? 15 : 3)) == 0;
> + // Shift into the range supported by Falk Hueffner's leap-year test:
> + // hueffner.de/falk/blog/a-leap-year-check-in-three-instructions.html
> + // Adding a multiple of 400 preserves divisibility by 4, 100, and 400.
> + // Idea by Cassio Neri: add 32800 (82 * 400).
> + // The conversion to uint32_t gives the algorithm's intended modulo 2^32
> + // arithmetic.
> + const auto __y = static_cast<uint32_t>(_M_y) + 32800u;
> + return ((__y * 1073750999u) & 3221352463u) <= 126976u;
> }
>
> explicit constexpr
> --
> 2.47.3
>
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