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~