lowbit is n & -n

Take a nonzero 32-bit int. From the low end, the first 1 together with the zeros below it is a power of two. That number is lowbit. For n = 0b110101000000, that is 0b000001000000.

First instinct

Walk bits until you find the 1.

int lowbit_naive(int n) {
  for (int k = 0; k <= 30; ++k) {
    if ((n & (1 << k)) > 0) {
      return 1 << k;
    }
  }
  return -1;
}

Flip, add one, AND

Then I flipped the word. The trailing zeros become ones; add one and they collapse back to zeros, and a one lands exactly where the original first 1 was. After that, n and ~n + 1 agree only on the lowbit — everything above it is opposite. Split as high bits / that 1 / the trailing zeros:

n           11010 1 000000
~n          00101 0 111111
~n + 1      00101 1 000000
n & (~n+1)  00000 1 000000

AND them together and you have it:

int lowbit_smart(int n) {
  return n & (~n + 1);
}

The pretty version

And ~n + 1 is just -n. So lowbit is n & -n.

int lowbit(int n) {
  return n & (-n);
}