[gcc r17-3387] libstdc++: Optimize chrono year::is_leap
Jonathan Wakely
redi@gcc.gnu.org
Tue Aug 18 18:54:45 GMT 2026
https://gcc.gnu.org/g:28b1b55c60464da23c65c6a8a9515c29f8d67ad8
commit r17-3387-g28b1b55c60464da23c65c6a8a9515c29f8d67ad8
Author: Francisco Muniz <munizfco@gmail.com>
Date: Wed Jul 29 21:57:10 2026 -0300
libstdc++: Optimize chrono year::is_leap
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.
Although the previous algorithm (by Cassio) has better throughput due to
better vectorization, we use this one to optimize for latency instead.
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>
Diff:
---
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 692fd6025e7d..4116f1c815eb 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
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