Use the & symbol to determine the parity of a number, and determine whether a number (x) is 2 of the n-th square

Judging the parity of a number is seen in C + + primer problem sets, and it feels interesting:

Suppose there is an integer x ha, assuming that

Then there are:

if (x&1)
    cout<< "odd" <<endl;
else
   cout<< "even" <<endl;

Take eight-bit binary as an example: 1 of the binary is 00000001 then an odd binary last must be 1, then an odd and 1 want to be 1, and conversely, even the last digit of the binary must be 0, so, an even number and a 1 want with a certain 0;

The return value of the following function (Microsoft) is calculated---------The number of statistics 1
int func (int x)
{
int countx = 0;
while (x)
{
countx++;
x = x& (x-1);
}
return countx;
}
Assume x = 9999
10011100001111
Answer: 8
Idea: Convert x into 2, and see the number of 1.
Note: each time x = x& (x-1) is executed, the rightmost 1 is changed to 0 when x is represented by binary, because x-1 will change the bit (x is the rightmost 1 in binary notation) to 0. (assuming that the bit is the K-bit, minus 1, after the bit all becomes 1, but because the bit is the rightmost 1, so the K-bit of x is all 0, and the K-bit of x-1 is 0, from 1 to K-1 is 1, and the result of the bitwise AND of the 1~K-1 bit is 0. ）
So one of the uses of the expression:

1: Turning the lowest bit of 1 into 0, this interpretation is not difficult to understand.

The second use of the expression is:
2: When x is odd,,x=x& (x-1) has the same value as x=x-1; when X is the N-Power of 2, the result is 0, which can be used to quickly determine whether a number is a 2 n-th square.
Determines whether a number (x) is 2 of the n-th square
#include <stdio.h>
int func (int x)
{
if ((x& (x-1)) = = 0)
return 1;
Else
return 0;
}

int main ()
{
int x = 8;
printf ("%d\n", func (x));
}
Note:
(1) If a number is a 2 of the N-square, then the number in binary representation of its highest bit is 1, the rest is 0, and then use the above idea is easy to think.