// <chrono> -*- C++ -*-
-// Copyright (C) 2008-2023 Free Software Foundation, Inc.
+// Copyright (C) 2008-2024 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library. This library is free
// software; you can redistribute it and/or modify it under the
# include <bits/unique_ptr.h>
#endif
+#define __glibcxx_want_chrono
+#define __glibcxx_want_chrono_udls
+#include <bits/version.h>
+
namespace std _GLIBCXX_VISIBILITY(default)
{
_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
static constexpr weekday
_S_from_days(const days& __d)
{
- auto __n = __d.count();
- return weekday(__n >= -4 ? (__n + 4) % 7 : (__n + 5) % 7 + 6);
+ using _Rep = days::rep;
+ using _URep = make_unsigned_t<_Rep>;
+ const auto __n = __d.count();
+ const auto __m = static_cast<_URep>(__n);
+
+ // 1970-01-01 (__n = 0, __m = 0 ) -> Thursday (4)
+ // 1969-31-12 (__n = -1, __m = _URep(-1)) -> Wednesday (3)
+ const auto __offset = __n >= 0 ? _URep(4) : 3 - _URep(-1) % 7 - 7;
+ return weekday((__m + __offset) % 7);
}
public:
friend constexpr days
operator-(const weekday& __x, const weekday& __y) noexcept
{
- auto __n = static_cast<long long>(__x._M_wd) - __y._M_wd;
- return days{__detail::__modulo(__n, 7)};
+ const auto __n = __x.c_encoding() - __y.c_encoding();
+ return static_cast<int>(__n) >= 0 ? days{__n} : days{__n + 7};
}
};
{
const auto __m = static_cast<unsigned>(month());
- // Excluding February, the last day of month __m is either 30 or 31 or,
- // in another words, it is 30 + b = 30 | b, where b is in {0, 1}.
+ // The result is unspecified if __m < 1 or __m > 12. Hence, assume
+ // 1 <= __m <= 12. For __m != 2, day() == 30 or day() == 31 or, in
+ // other words, day () == 30 | b, where b is in {0, 1}.
- // If __m in {1, 3, 4, 5, 6, 7}, then b is 1 if, and only if __m is odd.
- // Hence, b = __m & 1 = (__m ^ 0) & 1.
+ // If __m in {1, 3, 4, 5, 6, 7}, then b is 1 if, and only if, __m is
+ // odd. Hence, b = __m & 1 = (__m ^ 0) & 1.
- // If __m in {8, 9, 10, 11, 12}, then b is 1 if, and only if __m is even.
- // Hence, b = (__m ^ 1) & 1.
+ // If __m in {8, 9, 10, 11, 12}, then b is 1 if, and only if, __m is
+ // even. Hence, b = (__m ^ 1) & 1.
// Therefore, b = (__m ^ c) & 1, where c = 0, if __m < 8, or c = 1 if
// __m >= 8, that is, c = __m >> 3.
- // The above mathematically justifies this implementation whose
- // performance does not depend on look-up tables being on the L1 cache.
- return chrono::day{__m != 2 ? ((__m ^ (__m >> 3)) & 1) | 30
- : _M_y.is_leap() ? 29 : 28};
+ // Since 30 = (11110)_2 and __m <= 31 = (11111)_2, the "& 1" in b's
+ // calculation is unnecessary.
+
+ // The performance of this implementation does not depend on look-up
+ // tables being on the L1 cache.
+ return chrono::day{__m != 2 ? (__m ^ (__m >> 3)) | 30
+ : _M_y.is_leap() ? 29 : 28};
}
constexpr
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