Why is the battery life of the 1st parity log (n) long?

Question

So, I am reading the code from this site: http://www.geeksforgeeks.org/write-ac-program-to-find-the-parity-of-an-unsigned-integer/

And this shows how to determine if a number has even or odd parity. However, I do not understand why the execution efficiency is log (n). Here is the code for reference:

# include <stdio.h>
# define  bool int

/* Function to get parity of number n. It returns 1
   if n has odd parity, and returns 0 if n has even
   parity */
bool getParity(unsigned int n)
{
    bool parity = 0;
    while (n)
    {
        parity = !parity;
        n      = n & (n - 1);
    }        
    return parity;
}

Answer 1

Execution efficiency is O (log (n)), where nis the value of the integer that you pass. However, this is an unconventional way of using O notation.

, O ( , ), k = O (log2 (n)) O (k).

( , - Θ (s), s - , n, , - O (1)).

Answer 2

, 1, , , (1) (, p) n.

, Θ (p).

, n, log2 (n), , id O (log2 (n)).