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.
r~
