http://gcc.gnu.org/bugzilla/show_bug.cgi?id=50865
--- Comment #10 from joseph at codesourcery dot com <joseph at codesourcery dot com> 2011-10-25 17:13:51 UTC --- On Tue, 25 Oct 2011, jaak at ristioja dot ee wrote: > http://gcc.gnu.org/bugzilla/show_bug.cgi?id=50865 > > --- Comment #9 from Jaak Ristioja <jaak at ristioja dot ee> 2011-10-25 > 16:37:48 UTC --- > (In reply to comment #8) > > Well, they are equivalent where they are both defined, or if you apply C99 > > rules to infinite-precision integers. The problem here is that INT_MIN % > > -1 is undefined (explicitly in C1X) and so a transformation of INT_MIN % 1 > > into INT_MIN % -1 is unsafe (the other way round, transforming undefined > > behavior to defined, is fine at least in the absence of -ftrapv). > > But INT_MIN % 1 is still defined to be zero? Yes. INT_MIN % 1 is defined to be zero, since the infinite-precision values of both INT_MIN / 1 and INT_MIN % 1 are representable. INT_MIN % -1 is undefined because the infinite-precision value of INT_MIN / -1 is not representable in int.
