Le 24/02/2018 à 03:26, Richard Henderson a écrit :
> On 02/23/2018 06:59 AM, Laurent Vivier wrote:
>> @@ -4550,8 +4556,8 @@ int64_t floatx80_to_int64(floatx80 a, float_status
>> *status)
>> if ( shiftCount ) {
>> float_raise(float_flag_invalid, status);
>> if ( ! aSign
>> - || ( ( aExp == 0x7FFF )
>> - && ( aSig != LIT64( 0x8000000000000000 ) ) )
>> + || ((aExp == floatx80_infinity_high)
>> + && (aSig != floatx80_infinity_low))
>> ) {
>
> As long as you're cleaning this up, m68k ignores the explicit integer bit when
> considering an infinity. However, Intel doesn't ignore the bit -- it appears
> to treat 7fff.0* as a NaN.
According to the comment above the function:
"... If `a' is a NaN, the largest positive integer is returned.
Otherwise, if the conversion overflows, the largest integer with the
same sign as `a' is returned."
I think the change should be in fact:
@@ -4549,10 +4555,7 @@ int64_t floatx80_to_int64(floatx80 a,
float_status *status)
if ( shiftCount <= 0 ) {
if ( shiftCount ) {
float_raise(float_flag_invalid, status);
- if ( ! aSign
- || ( ( aExp == 0x7FFF )
- && ( aSig != LIT64( 0x8000000000000000 ) ) )
- ) {
+ if (!aSign || floatx80_is_any_nan(a)) {
return LIT64( 0x7FFFFFFFFFFFFFFF );
}
return (int64_t) LIT64( 0x8000000000000000 );
Do you agree?
Thanks,
Laurent
