libstdc++: Improve operator-(weekday x, weekday y)

The current implementation calls __detail::__modulo which is relatively
expensive.

A better implementation is possible if we assume that x.ok() && y.ok() == true,
so that n = x.c_encoding() - y.c_encoding() is in [-6, 6]. In this case, it
suffices to return n >= 0 ? n : n + 7.

The above is allowed by [time.cal.wd.nonmembers]/5: the returned value is
unspecified when x.ok() || y.ok() == false.

The assembly emitted for x86-64 and ARM can be seen in:
https://godbolt.org/z/nMdc5vv9n.

libstdc++-v3/ChangeLog:

	* include/std/chrono (operator-(const weekday&, const weekday&)):
	Optimize.

(cherry picked from commit f71352c71d78ac977ea0e71a6900699a8cf09219)

diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
index 0d3c081943947fda884c27e7e5ec12a4154f4f3d..3c5b425c7fe882f440ac4a4e6ac9845a3c6c6441 100644 (file)
--- a/libstdc++-v3/include/std/chrono
+++ b/libstdc++-v3/include/std/chrono
@@ -686,8 +686,8 @@ _GLIBCXX_BEGIN_NAMESPACE_VERSION
       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};
       }
 
       // TODO: operator<<, from_stream.