Bitwise Less than or Equal to

Question

There seems to be some kind of misconception that this is for a contest. I\'m trying to work through an assignment and I\'ve been stuck on this for an hour now.

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Answer 1

Really enjoyed Yanagar1's answer, which is very easy to understand.

Actually we can remove those shift operators and use De Morgan's laws, which reduce the number of operators from 15 to 11.

long isLessOrEqual(long x, long y) {
  long sign = (x ^ y);               // highest bit will be 1 if different sign
  long diff = sign & x;              // highest bit will be 1 if diff sign and neg x
  long same = sign | (y + (~x + 1)); // De Morgan's Law with the following ~same
                                     // highest bit will be 0 if same sign and y >= x
  long result = !!((diff | ~same) & 0x8000000000000000L); // take highest bit(sign) here
  return result;
}