This algorithm goes through as many iterations as there are set bits. So if we have a 32-bit word with only the high bit set, then it will only go once through the loop. In the worst case, it will pass once per bit. An integer n has log(n) bits, hence the worst case is O(log(n)). Here's your code annotated at the important bits (pun intended):
int count_set_bits(int n){ int count = 0; // count accumulates the total bits set while(n != 0){ n &= (n-1); // clear the least significant bit set count++; } }