On 06.02.19 04:08, Richard Henderson wrote:
> On 2/5/19 4:22 PM, David Hildenbrand wrote:
>> +static inline bool float128_is_normal(float128 a)
>> +{
>> + return ((a.high + (1ULL << 47)) & -1ULL >> 1) >= 1ULL << 48;
>
> I believe this is off by one: 1 << 48 and >= 1 << 49.
>
> The exponent is at [62:48]. The trick is adding 1, letting Inf+NaN overflow
> into the sign, masking out the sign, and checking that the result >= 2 to
> eliminate Inf+NaN (0) and Zero+Denormal (1).
Right, the exponent must not be 0 or all 1's. so after adding 1, it must
be >=2.
Yesterday it all made sense on my sheet of paper :) Let's verify against
float64
float64 has an exponent of 11:
return ((float64_val(a) + (1ULL << 52)) & -1ULL >> 1) >= 1ULL << 53;
float128 has an exponent of 15, the difference is 4. So 52 -> 48, 53 -> 49.
So you're right.
>
> I think this is clearer as
>
> (((a.high >> 48) + 1) & 0x7fff) >= 2.>
> It might be worth applying this to the other formats for clarity...
>
Yes, this makes it much clearer, will send a patch for the others as
well, thanks!
>
> r~
>
--
Thanks,
David / dhildenb
