>>> 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.
Oh, it needs to be absolutely portable? Well, you're already
assuming two's complement.
if (ux <= LONG_MAX) /* or whatever the type was */
x = ux;
else {
x = ux + LONG_MIN;
x += LONG_MIN;
}
should do the trick I think?
Segher
