3 ms·

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...