On Wed, Jan 12, 2011 at 03:05:10PM -0600, Peter Maydell wrote:
> On 12 January 2011 13:59, Aurelien Jarno <aurel...@aurel32.net> wrote:
>
> > @@ -494,7 +495,8 @@ int floatx80_is_quiet_nan( floatx80 a )
> > int floatx80_is_signaling_nan( floatx80 a )
> > {
> > #if SNAN_BIT_IS_ONE
> > - return ( ( a.high & 0x7FFF ) == 0x7FFF ) && (bits64) ( a.low<<1 );
> > + return ( ( a.high & 0x7FFF ) == 0x7FFF )
> > + && (LIT64( 0x8000000000000000 ) >= ((bits64) ( a.low<<1 )));
> > #else
> > bits64 aLow;
>
> If a is {0x7ffff,0} (ie +inf) this will return true, which is wrong.
> Do you want "<=" instead?
Correct, I swapped the operands at the last minute to match the other
functions, but without changing the sign.
> Actually, will
> return ((a.high & 0x7fff) == 0x7fff) && (a.low >=
> LIT64(0x4000000000000000));
> do? Untested but I think it will do the right thing. I'm not sure
The explicit bit might be one for a NaN, so you should filter it first.
> why this code has those bit64 casts, incidentally, since a.low is
> already a uint64_t.
Don't know either, but as they were already there, I left them.
> Also, maybe we should have a comment somewhere explaining
> why this is different from the other NaN functions (ie that the
> x80 format has an explicit bit and the others don't) ?
>
Good idea, will do.
--
Aurelien Jarno GPG: 1024D/F1BCDB73
aurel...@aurel32.net http://www.aurel32.net
