The current implementation returns
(_M_y & (__is_multiple_of_100 ? 15 : 3)) == 0;
where __is_multiple_of_100 is calculated using an obfuscated algorithm which
saves one ror instruction when compared to _M_y % 100 == 0 [1].
In leap years calculation, it's correct to replace the divisibility check by
100 with the one by 25. It turns out that _M_y % 25 == 0 also saves the ror
instruction [2]. Therefore, the obfuscation is not required.
[1] https://godbolt.org/z/5PaEv6a6b
[2] https://godbolt.org/z/55G8rn77e
libstdc++-v3/ChangeLog:
* include/std/chrono: Clear code.
---
Previous versions of this patch failed to apply. Hope this one works.
Good for trunk?
libstdc++-v3/include/std/chrono | 40 ++++++++++++++++-----------------
1 file changed, 20 insertions(+), 20 deletions(-)
diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
index 10e868e5a03..e18f57229e8 100644
--- a/libstdc++-v3/include/std/chrono
+++ b/libstdc++-v3/include/std/chrono
@@ -835,29 +835,29 @@ _GLIBCXX_BEGIN_NAMESPACE_VERSION
constexpr bool
is_leap() const noexcept
{
- // Testing divisibility by 100 first gives better performance, that is,
- // return (_M_y % 100 != 0 || _M_y % 400 == 0) && _M_y % 4 == 0;
-
- // It gets even faster if _M_y is in [-536870800, 536870999]
- // (which is the case here) and _M_y % 100 is replaced by
- // __is_multiple_of_100 below.
+ // 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://accu.org/journals/overload/28/155/overload155.pdf#page=16
-
- // Furthermore, if y%100 == 0, then y%400==0 is equivalent to y%16==0,
- // so we can simplify it to (!mult_100 && y % 4 == 0) || y % 16 == 0,
- // which is equivalent to (y & (mult_100 ? 15 : 3)) == 0.
- // See https://gcc.gnu.org/pipermail/libstdc++/2021-June/052815.html
-
- constexpr uint32_t __multiplier = 42949673;
- constexpr uint32_t __bound = 42949669;
- constexpr uint32_t __max_dividend = 1073741799;
- constexpr uint32_t __offset = __max_dividend / 2 / 100 * 100;
- const bool __is_multiple_of_100
- = __multiplier * (_M_y + __offset) < __bound;
- return (_M_y & (__is_multiple_of_100 ? 15 : 3)) == 0;
+ // [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;
}
explicit constexpr
--
2.41.0
