求两个数二进制中不同位的个数

方法一：

#include 
int count_diff_bit(int m, int n)
{
	int i = 0;
	int c = 0;//计数器
	for (i = 0; i < 32; i++)
	{
		if ((m & 1) != (n & 1))
		{
			c++;
		}
		m >>= 1;
		n >>= 1;
	}
	return c;
}
int main()
{
	int m = 0;
	int n = 0;
    scanf("%d %d",&m, &n);
	int len = count_diff_bit(m, n);
	printf("%d\n", len);
	return 0;
}

方法二：

#include 
int count_diff_bit(int m, int n)
{
	int c = 0;//计数器
	int tmp = m ^ n;//二进制的异或
	//统计tmp的二进制位中有多少个1:
	while (tmp)
	{
		tmp = tmp & (tmp - 1);
		c++;
	}
	return c;
}
int main()
{
	int m = 0;
	int n = 0;
    scanf("%d %d",&m, &n);
	int len = count_diff_bit(m, n);
	printf("%d\n", len);
	return 0;
}

或不写成函数形式：

#include 
int main()
{
    int m = 0;
    int n = 0;
    scanf("%d %d", &m, &n);
    int c = 0;
    int tmp = m ^ n;
    while (tmp)
    {
        tmp = tmp & (tmp - 1);
        c++;
    }
    printf("%d\n", c);
    return 0;
}