On 23/06/21 18:51 +0100, Jonathan Wakely wrote:
Here's what I've committed. Tested x86_64-linux and powerpc64le-linux.
Pushed to trunk.
commit b92d12d3fe3f1aa56d190d960e40c62869a6cfbb
Author: Cassio Neri <cassio.n...@gmail.com>
Date: Wed Jun 23 15:32:16 2021
libstdc++: More efficient std::chrono::year::leap
Simple change to std::chrono::year::is_leap. If a year is multiple of 100,
then it's divisible by 400 if and only if it's divisible by 16. The latter
allows for better code generation.
The expression is then either y%16 or y%4 which are both powers of two
and so it can be rearranged to use simple bitmask operations.
Co-authored-by: Jonathan Wakely <jwak...@redhat.com>
Co-authored-by: Ulrich Drepper <drep...@redhat.com>
libstdc++-v3/ChangeLog:
* include/std/chrono (chrono::year::is_leap()): Optimize.
diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
index 4631a727d73..863b6a27bdf 100644
--- a/libstdc++-v3/include/std/chrono
+++ b/libstdc++-v3/include/std/chrono
@@ -1606,13 +1606,18 @@ _GLIBCXX_BEGIN_NAMESPACE_VERSION
// [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 rearrange the expression to (mult_100 ? y % 4 : y % 16)==0
But Ulrich pointed out I got my boolean logic all muddled up in the
comment. Fixed with the attached patch!
commit 4a404f66b09d661bdc60e7e0d5d08f00c4e194db
Author: Jonathan Wakely <jwak...@redhat.com>
Date: Wed Jun 23 18:50:03 2021
libstdc++: Fix comment in chrono::year::is_leap()
libstdc++-v3/ChangeLog:
* include/std/chrono (chrono::year::is_leap()): Fix incorrect
logic in comment.
diff --git a/libstdc++-v3/include/std/chrono b/libstdc++-v3/include/std/chrono
index 863b6a27bdf..0b597be1944 100644
--- a/libstdc++-v3/include/std/chrono
+++ b/libstdc++-v3/include/std/chrono
@@ -1606,8 +1606,8 @@ _GLIBCXX_BEGIN_NAMESPACE_VERSION
// [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 rearrange the expression to (mult_100 ? y % 4 : y % 16)==0
+ // 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