https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #9 from Iain Buclaw ---
Another related issue has been created in pr97889.
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
Iain Buclaw changed:
What|Removed |Added
Status|UNCONFIRMED |RESOLVED
Resolution|---
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #7 from CVS Commits ---
The releases/gcc-10 branch has been updated by Iain Buclaw
:
https://gcc.gnu.org/g:bbb887834d78cf6a444bf9cecc29d14b4dfb9cf8
commit r10-9043-gbbb887834d78cf6a444bf9cecc29d14b4dfb9cf8
Author: Iain Buclaw
Date:
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #6 from Alex ---
>From cppreference.com :
The behavior of every builtin compound-assignment expression E1 op= E2 (where
E1 is a modifiable lvalue expression and E2 is an rvalue expression or a
braced-init-list (since C++11)) is exact
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #5 from Iain Buclaw ---
(In reply to Alex from comment #2)
> The arithmetic equivalent would be for:
> X += 4/2
> To be produce:
> Immediate load Register with 4
> Add register with 4 in it to x
> Divide register with 4 in it by 2
> R
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #4 from Iain Buclaw ---
Might be a regression caused by the fix for PR96924
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #3 from Iain Buclaw ---
(In reply to Alex from comment #2)
> I agree that the order of evaluation of operands is undefined and writing
> code that depends on that order would not be reliable. In this case it's the
> execution of the a
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #2 from Alex ---
I agree that the order of evaluation of operands is undefined and writing code
that depends on that order would not be reliable. In this case it's the
execution of the assign expression happening before all the operan
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=97843
--- Comment #1 from Iain Buclaw ---
(In reply to Alex from comment #0)
> This code:
>
> import std.stdio;
>
> ubyte sum(ubyte[] bytes)
> {
> ubyte result;
> foreach(x;bytes)
> result += x;
> return result;
> }
>
