Calculate n + 1 without using + or - or * or /

    def increment(a):
        import urllib2
        url = "http://www.html2xml.nl/Services/Calculator/Version1/Calculator.asmx/Add?a=%d&b=1" % a       
        result = urllib2.urlopen(url).read() 
        from xml.dom.minidom import parseString
        result = parseString(result)
        return int(result.getElementsByTagName('int')[0].childNodes[0].data)

    def inc(n):
        x = 1
        while n & 1 == 1:
            n = n >> 1
            x = x << 1
        n = n | 1
        while not x == 1:
            x = x >> 1
            n = n << 1
        return n

  int inc(int a) {
    int mask = ~a;
    /* Assuming a is 32-bit. If 64-bit, add 32-bit shift. */
    mask |= mask << 1;
    mask |= mask << 2;
    mask |= mask << 4;
    mask |= mask << 8;
    mask |= mask << 16;
    /* Take the longest run of 1s from the right of a.
     * mask has 0s there and 1s everywhere else. We want
     * to set the next bit to 1 and set the ones after
     * to 0. */
    return (a | ~(mask << 1)) & mask;
  }