4 ms·

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 optimizab

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 :)