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2^-infinity == 0


And 3 == 2^(log_2(3)).

Clearly we want to know if x is an integer power of 2.


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.


Actually, it's undefined behaviour. http://blog.regehr.org/archives/213


2^(bitsize) == 2^(bitsize)

If you meant 1 << (bitsize), that's undefined behaviour in C




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