[llvm-commits] PATCH: Preserving the 'nsw' flag in the instruction combiner.

[Nick Lewycky]

On 10 August 2011 13:19, Eli Friedman <eli.friedman at gmail.com> wrote:

> On Wed, Aug 10, 2011 at 1:11 PM, Nick Lewycky <nlewycky at google.com> wrote:
> > On 10 August 2011 13:07, Eli Friedman <eli.friedman at gmail.com> wrote:
> >>
> >> On Wed, Aug 10, 2011 at 12:57 PM, Nick Lewycky <nlewycky at google.com>
> >> wrote:
> >> > On 10 August 2011 12:37, Pranav Bhandarkar <pranavb at codeaurora.org>
> >> > wrote:
> >> >>
> >> >> Hi Nick,
> >> >>
> >> >> >>Thanks for working on this! Firstly, I have a high-level question:
> >> >> >> why
> >> >> can't (X +nsw (C1 +nsw C2) always become (X +nsw C3)? Your patch
> spends
> >> >> a
> >> >> lot of time verifying that >>overflow couldn't occur, but you're
> given
> >> >> an
> >> >> assumption a priori that it can't because the nsw flag is present.
> >> >>
> >> >> The reason is that 127 (C1) + 2 (C2) = 129 (C3) is not strictly true
> >> >> always.
> >> >> For instance, when the type is i8, then C3 is -127.
> >> >
> >> > 127 + 2 = 129 remains true with i8 values, because i8 -127 = i8 129.
> >> > They
> >> > have the same bit pattern.
> >> > Can you pick values for X, C1 and C2 which still satisfy the +nsw
> >> > relationship between each other that can't be converted into X +nsw C3
> >> > where
> >> > C3 = C1+C2 (no nsw requirement)? 127 and 2 don't work as
> >> > counter-examples
> >> > because that triggers undefined behaviour (it violates the 'nsw' bit
> >> > present
> >> > on the add).
> >>
> >> X = -128, C1 = 127, C1 = 2?
> >
> > I assume C2 = 2 and all i8 values. That doesn't work because the
> expression
> > was "(X +nsw (C1 +nsw C2)" and 127 +nsw 2 invokes undefined behaviour,
> > crossing from 127 to -128/128 to reach -127/129. In fact nsw-adding
> anything
> > positive to 127 is invalid, so you'd have to be sure to pick a negative
> C2
> > given a C1 of 127.
>
> The original expression is ((X +nsw C1) +nsw C2).
>

Doh! Thanks.

So the original code with your values does:
((X +nsw C1) +nsw C2) --> ((-128 +nsw 127) +nsw 2) = 1, and none of the
nsw's are violated. Good.

After my proposed transform, we get:
C3 = C1 + C2 --> C3 = 127 + 2 = -127.
(X +nsw C3) --> (-128 +nsw -127) = 1, and the nsw wasn't violated (we moved
towards zero 129 times, starting from -128 up past -1 and crossed the
unsigned boundary to 0 finally landing at 1, never crossing the signed
boundary).

I may still be wrong, but I don't think so. :-) More generally I think we
can always take the intersection of the nuw/nsw/exact flags available on any
string of the same instruction. Start mixing instructions and things get
messy though.

Nick
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.llvm.org/pipermail/llvm-commits/attachments/20110810/036c3209/attachment.html>