Florian Weimer wrote: > * Andrew Haley: > >> The test was, if I recall correctly >> >> x = a + b; >> if ((x ^ a) & (x ^ b)) < 0) >> >> all you have to do is convert everything to unsigned values, then >> >> ux = ua + ub; >> if ((ux ^ ua) & (ux ^ ub)) & (unsigned)INT_MIN)) >> goto deal_with_overflow; >> // we now know there is no overflow >> x = ux; >> >> which is exactly the same test as before, but perfectly compliant. > > The comment is wrong. The code checks for signed overflow, but the > following assignment still overflwos when ux is larger than INT_MAX. > So this version is usable exactly under the same circumstances as the > first one. Ahhh, I see. Hmm, there must be a decent way to do this. > And the comparison against 0 results in tighter code. 8-) Heh. Andrew.
