6 ms·

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