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Date: Wed, 20 Apr 2011 11:35:51 +0100
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From: Peter Maydell <peter.maydell@linaro.org>
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Subject: Re: [Qemu-devel] [PATCH v2 01/20] softfloat: fix floatx80 handling
	of NaN
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To: Aurelien Jarno <aurelien@aurel32.net>
Cc: qemu-devel@nongnu.org

On 20 April 2011 11:11, Aurelien Jarno <aurelien@aurel32.net> wrote:
> @@ -624,10 +630,11 @@ static floatx80 commonNaNToFloatx80( commonNaNT a S=
TATUS_PARAM)
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 return z;
> =C2=A0 =C2=A0 }
>
> - =C2=A0 =C2=A0if (a.high)
> - =C2=A0 =C2=A0 =C2=A0 =C2=A0z.low =3D a.high;
> - =C2=A0 =C2=A0else
> + =C2=A0 =C2=A0if (a.high >> 1) {
> + =C2=A0 =C2=A0 =C2=A0 =C2=A0z.low =3D LIT64( 0x8000000000000000 ) | a.hi=
gh >> 1;
> + =C2=A0 =C2=A0} else {
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 z.low =3D floatx80_default_nan_low;
> + =C2=A0 =C2=A0}
> =C2=A0 =C2=A0 z.high =3D ( ( (uint16_t) a.sign )<<15 ) | 0x7FFF;
> =C2=A0 =C2=A0 return z;
> =C2=A0}

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