Count the number of 1 in the binary

This article provides three ways to calculate the number of 1 in a binary representation of a number, respectively. Methods and explanations are shown in Count1, Count2, Count3 function respectively.

Only Count1 can not meet the negative demand (will die loop in), the other two will be able to operate within 32b positive negative numbers.

Count1: Each time x is last to 1, see if it is 1, then move X to the right

Count2: Start the variable y from 1 to X, and then move Y to the left each time, similar to the previous method

Count3: Each time x&= (x-1) can change the rightmost one 1 to 0;

The method is applied: e.g. how to use a statement to determine whether an integer is 2 of the power of the whole number?

/********************///@Discription: Count The number of ' 1 ' in binary number//@Author: Rachel Zhang//@Create date:2012-5-22/********************/#include"stdio.h"intCount1 (intx) {    intC=0;  while(x) {C+=x&1; X>>=1; }    returnC;}//Bug: When x=0x80000000 (negative), x>>1 to guarantee or negative, so not 0x40000000, but =0xc0000000//i.e input-2147483648 (0x80000000), x>>1=-107374124 (0xc0000000), not 2^30 (0x40000000)//so the final 0xff will lead to a dead loop .intCount2 (intx) {    intC=0; unsignedinty=1;  while(y) {c+=x&y?1:0; Y<<=1; }    returnC;}//problem: Do not know the number of x digits, will cause some redundant operations, but it's okay, 32-bit PC up to 32 times//It's not a problem.intCount3 (intx) {    intC=0;  while(x) {x&=x-1; C++; }    returnC;}intMain () {intx;  while(SCANF ("%d", &x)! =EOF) {printf ("%d\n%d\n%d\n", Count1 (x), Count2 (x), Count3 (x)); }    return 0;}

SOURCE title: http://zhedahht.blog.163.com/blog/static/25411174200731844235261/

Count the number of 1 in the binary