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

[Ulf Magnusson]

On Thu, Oct 6, 2011 at 11:04 AM, Miles Bader <miles@gnu.org> wrote:
> Ulf Magnusson <ulfalizer@gmail.com> writes:
>> Might as well do
>>
>> bool overflowbit(unsigned int a, unsigned int b) {
>>     const unsigned int sum = a + b;
>>     return (a ^ b) & ~(a ^ sum) & 0x80;
>> }
>>
>> But still not very good output compared to other approaches as expected.
>
> How about:
>
>   bool overflowbit2(unsigned int a, unsigned int b)
>   {
>       const unsigned int sum = a + b;
>       return ~(a ^ b) & sum & 0x80;
>   }
>
> ?
>
> I thinnnnk it has the same results as your function...
> [I just made a table of all 8 possibilities, and checked!]
>
> -miles
>
> --
> Circus, n. A place where horses, ponies and elephants are permitted to see
> men, women and children acting the fool.
>

Ops, should have been

return ~(a ^ b) & (a ^ sum) & 0x80

~(a ^ b) gives 1 in the sign bit position if the signs are the same,
and (a ^ sum) gives 1 if it's different in the sum.

A clearer way of writing it (that also generates suboptimal code) is

bool overflow(unsigned int a, unsigned int b) {
    const unsigned asign   = a       & 0x80;
    const unsigned bsign   = b       & 0x80;
    const unsigned sumsign = (a + b) & 0x80;
    return (asign == bsign) && (asign != sumsign);
}

Seems bit-fiddling isn't the way to go.

Maybe I should take this to gnu-help as it isn't really development-related.

/Ulf