46 ms·

One they missed: (x & -x) == x

• Jabbles 15y ago
  Unfortunately this doesn't work for x=0.
  • rcfox 15y ago
    2^-infinity == 0
    • Jabbles 15y ago
      And 3 == 2^(log_2(3)). Clearly we want to know if x is an integer power of 2.
      • wnoise 15y ago
        2^(bitsize) == 0. Edit: Yes, 1 << bitsize is undefined. But unsigned integers actually do have well-defined semantics on overflow, and multiplying by 2 enough times really does produce zero.
        • rcfox 15y ago
          Actually, it's undefined behaviour. http://blog.regehr.org/archives/213 http://blog.regehr.org/archives/213
        • ot 15y ago
          2^(bitsize) == 2^(bitsize) If you meant 1 << (bitsize), that's undefined behaviour in C
  • mansr 15y ago
    So make it x && blah. Do I really have to state the obvious?
    • Jabbles 15y ago
      You're a computer programmer (probably). Yes you have to state the obvious...
• chime 15y ago
  That was #9: int isPowerOfTwo (unsigned int x) { return ((x != 0) && !(x & (x - 1))); }
• quacker 15y ago
  This is method 10 in the article, since -x is equivalent to ~x+1.
• cube13 15y ago
  This works for negative values of x, though. It should be: ((x>0) && ((x & -x) == x))