3 ms·

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

by 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 :)