写一个函数返回参数二进制中 1 的个数

1.#include<stdio.h>

int bit_count(unsigned int n)

{

    int count;

    for(count = 0; n; n=n &(n - 1))

    {

        count++;

    }

    return count;

}

int main()

{   int k;

    int num;

printf("输入一个数：");

    scanf("%d",&num);

k=bit_count(num);

printf("1的个数为%d\n",k);

return 0;

}

2.

#include <stdio.h>

int main()

{

int count = 0;

int num = 0;

scanf("%d",&num);

while(num)

{

count++;

num = num & (num-1);

}

printf("count = %d\n",count);

return 0;

}