On 6/5/20 12:01 PM, Joseph Myers wrote:
> The floatx80 remainder implementation unnecessarily sets the high bit
> of bSig explicitly. By that point in the function, arguments that are
> invalid, zero, infinity or NaN have already been handled and
> subnormals have been through
The floatx80 remainder implementation unnecessarily sets the high bit
of bSig explicitly. By that point in the function, arguments that are
invalid, zero, infinity or NaN have already been handled and
subnormals have been through normalizeFloatx80Subnormal, so the high
bit will already be set.
