By using the & (logical and) operator, we can easily (and quickly!) test whether an integer is a power of 2.

In C:

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

This works because an integer that is a power of two, x, is represented by a single 1 bit, with the other bits being 0. The same bit in x-1 will always be 0 because of the subtraction. Thus, when the two are anded together, the result will always be 0.

For example:

- x = 1: x = 0001, x-1 = 0000, x & x-1 = 0001 & 0000 = 0000
- x = 2: x = 0010, x-1 = 0001, x & x-1 = 0010 & 0001 = 0000
- x = 4: x = 0100, x-1 = 0011, x & x-1 = 0100 & 0011 = 0000
- x = 8: x = 1000, x-1 = 0111, x & x-1 = 1000 & 0111 = 0000

But if x is not a power of two, the result is not 0:

- y = 3: y = 0011, y-1 = 0010, y & y-1 = 0011 & 0010 = 0010
- y = 5: y = 0101, y-1 = 0100, y & y-1 = 0101 & 0100 = 0100

**Update:**

I have come across another way to quickly test if an integer is a power of two.

In C:

```
int isPowerOfTwo(unsigned int x)
{
return ((x != 0) && ((x & (~x + 1)) == x));
}
```

See if you can figure this one out for yourself.