On 4/5/08, Andrew Haley <aph@xxxxxxxxxx> wrote:
> Tim Prince wrote:
> > Alexander Monakov wrote:
> >>> we are currently investigating some numerical algorithms and the claim
> >>> appeared that a C statement like
> >>> x = c - (c - a);
> >>> would be easily transformed into
> >>> x = a;
> >>> by the compiler. Now investigating this with a vanilla GCC 4.1.2 failed
> >>> to support the claim. Compiling the below program with -O3 -ffast-math
> >>> keeps the computation of x. The output shows x is different from a. The
> >>> question is, is there some compiler switch or the like to get GCC to
> >>> make the above transformation? I searched the docs but had the
> >>> impression that all relevant flags should be included in the above two
> >>> (especially ffast-math).
> >>
> >> This transformation is indeed included into -ffast-math. I checked with
> >>
> >> $ gcc --version
> >> gcc (GCC) 4.1.1 20070105 (Red Hat 4.1.1-52)
> >>
> >> and it does eliminate the calculation. What does generated assembly
> >> code look
> >> like in your case? Note you may as well check on this code:
> >>
> >> double f(double a, double c)
> >> {
> >>
> > Normally, when such expressions are written in source code, there is a
> > reasonable expectation that algebraic simplification will not be
> > performed across parentheses.
>
> There may well be such an expectation, but gcc only prevents re-association
> across explicit parenthesis in FORTRAN, and it's only been doing that for
> a few weeks.
Wait, so are things like "c - (c - a)" optimzed down to "a" or not? I
use a lot of that in very time-intensive c++ code.
