https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
--- Comment #10 from CVS Commits ---
The master branch has been updated by Jason Merrill :
https://gcc.gnu.org/g:df118d7ba138cacb17203d4a1b5f27730347cc77
commit r13-1783-gdf118d7ba138cacb17203d4a1b5f27730347cc77
Author: Jason Merrill
Date: W
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
Jonathan Wakely changed:
What|Removed |Added
Target Milestone|--- |11.0
Status|NEW
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
--- Comment #8 from CVS Commits ---
The master branch has been updated by Jonathan Wakely :
https://gcc.gnu.org/g:3ee44d4c518d61c6bbf75fcf280edc6ce5326ce0
commit r11-3759-g3ee44d4c518d61c6bbf75fcf280edc6ce5326ce0
Author: Jonathan Wakely
Date:
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
--- Comment #7 from Jonathan Wakely ---
Despite the code being correct, I think it would be better to hoist this branch
out of the loop:
if (__k == 0)
__r2 += __s;
else if (__k <= __s)
__r2 += __kn + _
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
Nathan Sidwell changed:
What|Removed |Added
Status|RESOLVED|NEW
Last reconfirmed|
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
--- Comment #5 from AK ---
> So when __k == 0, then all three of those loads will be _Type(0x8b8b8b8bu)
> really; no matter what the values of __n, __p will be.
Will it be a good idea to add the explanation in comments, as this may be
tricky f
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
--- Comment #4 from AK ---
Makes sense. Thanks for the explanation.
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
--- Comment #3 from Andrew Pinski ---
>In particular when __k is zero we're taking '(2^64 - 1) mod __n', which is not
>necessarily __n - 1, the last position in the buffer, right?
Yes that is correct except when __k is 0, then __begin will the
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
Andrew Pinski changed:
What|Removed |Added
Resolution|--- |WONTFIX
Status|UNCONFIRMED
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94823
AK changed:
What|Removed |Added
CC||hiraditya at msn dot com
--- Comment #1 from AK --
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