Option to make unsigned->signed conversion always well-defined?

[Pedro Pedruzzi]

Em 07-10-2011 02:35, Miles Bader escreveu:
> Pedro Pedruzzi <pedro.pedruzzi@gmail.com> writes:
>> On Thu, Oct 6, 2011 at 11:04 AM, Miles Bader <miles@gnu.org> wrote:
>>> How about:
>>>
>>>   bool overflowbit2(unsigned int a, unsigned int b)
>>>   {
>>>       const unsigned int sum = a + b;
>>>       return ~(a ^ b) & sum & 0x80;
>>>   }
>>
>> Miles, it is not the same. Take for example (0xff, 0xff). In 8-bit
>> 2's complement, this is (-1, -1) and does not overflow. Your
>> function says it does.
> 
> Negative overflow isn't considered overflow...?  wacky...

It is. For example -100 + -100 = -200 (less than INT8_MIN; does not
fit). But -1 + -1 = -2, is ok.

-- 
Pedro