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.

diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
index a826982803b7855bfdd6d8ce15b2be677c621efa..45e7d269e4b41a8f340e429357e805b1f2b337d4 100644 (file)
--- 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