4 ms·

> for example, that (x&~1*2)/2 == x Fixed that for you.

• caf 8y ago
  No? If x is 3, then (x * 2) / 2 is also equal to 3 (as per the GP) but (x & ~1 * 2) / 2 is equal to 0. (If you meant ((x & ~1) * 2) / 2, then that is equal to 2).
  • cornholio 8y ago
    You are of course correct. Not only have I mistook precedence, but I confused the lower bit with the upper one where overflow occurs. So the compiler optimizable fix is something like: ((x & INT_MAX) * 2) / 2 = x
    • caf 8y ago
      ((x & (INT_MAX / 2)) * 2) / 2 ?
      • cornholio 8y ago
        My version works up to x = INT_MAX and your version still fails for negative ints, so I don't see the benefit of restricting the range. Getting closer :)