http://gcc.gnu.org/bugzilla/show_bug.cgi?id=60578
--- Comment #5 from Zhendong Su <su at cs dot ucdavis.edu> --- (In reply to Jakub Jelinek from comment #4) > But why do you think (c ^ 0L) < -1 is 1 for 64-bit mode? > From C99, 6.3.1.8, different paragraphs apply: > "Otherwise, if the type of the operand with signed integer type can represent > all of the values of the type of the operand with unsigned integer type, then > the operand with unsigned integer type is converted to the type of the > operand with signed integer type. > > Otherwise, both operands are converted to the unsigned integer type > corresponding to the type of the operand with signed integer type." > > For 64-bit mode, the first cited paragraph applies, because 64-bit long int > can > represent all of values of 32-bit unsigned int, so it is > (((long int) 0) ^ 0L) < -1L and thus 0L < -1L and therefore false. > > For 32-bit mode, the second cited paragraph applies, because 32-bit long int > can't represent all of values of 32-bit unsigned int, so it is > (((unsigned long int) 0) ^ 0UL) < -1UL and thus 0UL < -1UL and therefore > true. Thanks Jakub! Sorry for the invalid report.
