https://gcc.gnu.org/bugzilla/show_bug.cgi?id=45464
--- Comment #3 from Andrew Pinski <pinskia at gcc dot gnu.org> --- Actually I think only the warning for the three add case is wrong and here is why: adding two (unsigned) 8bit integers together will only give you a max a 9bit integer. so comparing b and (a+a) is fine Adding three 8bit (unsigned) integers together will give a 10bit integer. It just happens GCC can figure out the first case of adding two 8bit integers will not change value for signed vs unsigned while adding the third add, for the warning, gcc does not figure that out.
