二进制中1的个数

//第一种：短除法
int count_one(unsigned num)
{
 int count = 0;
 while (num != 0)
 {
  if (num % 2 == 1)
  {
count++;
  }
  num /= 2;
 }
 return count;
}
?
//第二种：移位按位与1法
int count_one(int num)
{
 int count = 0;
 int i = 0;
 for (i = 0; i < 32; i++)
 {
  count += (num >> i) & 1;
 }
 return count;
}
?
//第三种：num与num-1进行按位与
int count_one(int num)
{
 int count = 0;
 while (num != 0)
 {
  num = num & (num - 1);
  count++;
 }
 return count;
}
?
int main()
{
 int num = 0;
 int count = 0;
 printf("请输入一个数字>:");
 scanf("%d", &num);
 count = count_one(num);
 printf("1的个数为:%d\n", count);
 return 0;
}