写一个函数返回参数二进制中 1 的个数

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写一个函数返回参数二进制中 1 的个数(1)#include <stdio.h>int main(){int num = 0;int i = 0;int count = 0;        scanf("%d",&num);for(i=0; i<32; i++){            if((num&1) == 1){count++;}num = num>>1;/*if(((num>>i)&1) == 1)//任何一个数字，给它按位与（&）"1"，就能得知其最低位。{count++;}*/}printf("%d\n", count);return 0;}(2)#include <stdio.h>int main(){int num = 0;int count = 0;scanf("%d",&num);while(num){count++;num = num&(num-1);//一个数和这个数减一，按位与(&)，可以去掉最右面的"1"。}printf("count = %d\n", count);return 0;}(3)#include <stdio.h>int count_one_bits(unsigned int value){int count = 0;while(value){if(value%2 == 1){count++;}value = value/2;}return count;}int main(){unsigned int num = 0;int ret = 0;        printf("请输入一个数:\n");scanf("%d",&num);ret = count_one_bits(num);printf("二进制中1的个数为%d\n",ret);return 0;}

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