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libstdc++: Simplify year::is_leap() 

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 (year::is_leap): Clear code.
This commit is contained in:
Cassio Neri 2023-11-11 22:59:50 +00:00 committed by Jonathan Wakely
parent b011535456
commit 86a0df1a6c
1 changed files with 19 additions and 19 deletions
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38
libstdc++-v3/include/std/chrono
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@ -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
// [2] https://godbolt.org/z/55G8rn77e
// [3] https://gcc.gnu.org/pipermail/libstdc++/2021-June/052815.html
// 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;
return (_M_y & (_M_y % 25 == 0 ? 15 : 3)) == 0;
}
explicit constexpr