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

So make it x && blah. Do I really have to state the obvious?

You're a computer programmer (probably). Yes you have to state the obvious...