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;
}