On 20 April 2011 11:11, Aurelien Jarno <aurel...@aurel32.net> wrote:
> @@ -624,10 +630,11 @@ static floatx80 commonNaNToFloatx80( commonNaNT a
> STATUS_PARAM)
> return z;
> }
>
> - if (a.high)
> - z.low = a.high;
> - else
> + if (a.high >> 1) {
> + z.low = LIT64( 0x8000000000000000 ) | a.high >> 1;
> + } else {
> z.low = floatx80_default_nan_low;
> + }
> z.high = ( ( (uint16_t) a.sign )<<15 ) | 0x7FFF;
> return z;
> }
This is still retaining the sign bit from the input if it generates
a default NaN because the mantissa would have been zero. This isn't
consistent with the commonNaNToFloat64/32, which just return the
float64/32_default_nan with whatever sign it has.
-- PMM
