Determine the sign of a 32 bit int

Question

Using ONLY:

! ~ & ^ | + << >>

NO LOOPS

I need to determine the sign of a 32 bit integer and I need to return 1 if positive, 0 if 0 and -1 if ne

Answer 1

Dimitri's idea could be simplified to (!!x) - ((x >> 30) & 2)

And just to give one more cryptic solution:

~!x & ((-((unsigned) x >> 31)) | !!x)

Answer 2

What about:

int getsign(int n)
{
  return (!!n) + (~((n >> 30) & 2) + 1);
}

..for 32-bit signed int, 2's complement only.

!!n gives 1 if n is nonzero. ((n >> 30) & 2) gives 2 iff the high bit (sign) is set. The bitwise NOT and +1 take the 2's complement of this, giving -2 or 0. Adding gives -1 (1 + -2) for negative values, 0 (0 + 0) for zero, and +1 (1 + 0) for positive values.