------- Comment #3 from segher at kernel dot crashing dot org 2008-02-26 16:16
-------
> > Not equivalent in the presence of overflow.
>
> You mean defined overflow :).
No, I mean overflow.
Let's assume int is 16-bit (just to keep the numbers smallish);
now take i=1, j=1000, k=1000. i/j/k is perfectly well-defined,
but i/(j*k) overflows in the j*k computation.
Segher
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segher at kernel dot crashing dot org changed:
What |Removed |Added
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CC| |segher at kernel dot
| |crashing dot org
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=35307
