https://gcc.gnu.org/bugzilla/show_bug.cgi?id=95650
Wilco <wilco at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Last reconfirmed| |2020-06-12
Ever confirmed|0 |1
CC| |wilco at gcc dot gnu.org
Status|UNCONFIRMED |NEW
--- Comment #4 from Wilco <wilco at gcc dot gnu.org> ---
(In reply to Alex Coplan from comment #3)
> I think clang's optimisation is sound here.
>
> C says that we add two shorts as int and then truncate to short (i.e. reduce
> mod 16).
>
> The question is whether the top bits being set (which the ABI allows) can
> influence the result. I don't think it can.
>
> The observation is that the "top bits being set" are just extra multiples of
> 2^16 in the addition, which just disappear when we reduce mod 2^16. That is:
>
> (x_1 + x_2 + y_1 + y_2) % 2^16 = (x_1 + x_2) % 2^16
>
> where x_1,x_2 are arbitrary integers and y_1,y_2 are multiples of 2^16 (the
> top bits).
Confirmed. It works for signed as well and any operator except right shift and
division. Basically the store requires only the bottom 16 bits to be valid, and
a backwards dataflow can propagate this to remove unnecessary zero and sign
extends.
