That was #9:

    int isPowerOfTwo (unsigned int x)
    {
      return ((x != 0) && !(x & (x - 1)));
    }

This works for negative values of x, though.

It should be: ((x>0) && ((x & -x) == x))

This is method 10 in the article, since -x is equivalent to ~x+1.

Unfortunately this doesn't work for x=0.

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...