Designing function f(f(n)) == -n

Question

A question I got on my last interview:

Design a function f, such that:

f(f(n)) == -n

Where n<

Answer 1

Depending on your platform, some languages allow you to keep state in the function. VB.Net, for example:

Function f(ByVal n As Integer) As Integer
    Static flag As Integer = -1
    flag *= -1

    Return n * flag
End Function

IIRC, C++ allowed this as well. I suspect they're looking for a different solution though.

Another idea is that since they didn't define the result of the first call to the function you could use odd/evenness to control whether to invert the sign:

int f(int n)
{
   int sign = n>=0?1:-1;
   if (abs(n)%2 == 0)
      return ((abs(n)+1)*sign * -1;
   else
      return (abs(n)-1)*sign;
}

Add one to the magnitude of all even numbers, subtract one from the magnitude of all odd numbers. The result of two calls has the same magnitude, but the one call where it's even we swap the sign. There are some cases where this won't work (-1, max or min int), but it works a lot better than anything else suggested so far.