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According to Bruno Haible on 3/25/2007 9:57 AM:
> Eric Blake wrote:
>> Is this any more portable, by avoiding floating point division altogether?
>>
>> /* return true iff the representation of d needs a leading '-' */
>> bool
>> is_negative (long double d)
>> {
>> if (d == 0)
>> {
>> union {
>> long double d;
>> long l;
>> } u;
>> u.d = d;
>> u.l |= 1;
>> return u.d < 0;
>> }
>> return d < 0;
>> }
>
> You are picking a particular bit in a 'long double' representation. If you
> picked any bit different from the sign bit, this code is fine (assuming
> !isnanl(d) is already known). If you picked the sign bit, +0.0 will be
> considered negative too.
Actually, if I picked the sign bit, then both -0 and +0 would be
considered positive, since neither 0 compares less than 0.
> It's an endianness issue and depends on the bit
> storage order in words. You might have picked the wrong bit for m68k...
> It's safer to use u.l |= 8; since noone will put the sign bit at
> bit 3 or 28 (except the HP-PA designers perhaps :-)).
Yes, I like your idea of using a bit other than the LSB as a way to ensure
that the sign bit is not hit. But you also have to pick a bit that does
not fall in the 2-byte padding of the 10-byte x86 long double.
- --
Don't work too hard, make some time for fun as well!
Eric Blake [EMAIL PROTECTED]
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