//1.写一个函数返回参数二进制中 1 的个数
#include<stdio.h>
int div = 0;   //除数
int rem = 0;   //余数
int count = 0; //计1
int count_one_bits(unsigned int div)
{int con = 0;   //商while (div >= 1){con = div / 2;rem = div % 2;div = con;if (1 == rem){++count;}}return 0;
}
int main()
{printf("请输入一个数字");scanf("%d", &div);count_one_bits(div);printf("%d", count);return 0;
}//2.获取一个数二进制序列中所有的偶数位和奇数位，分别输出二进制序列。
#include<stdio.h>
int main()
{int i = 0;int j = 0;int num = 0;int arr1[16] = { 0 };int arr2[16] = { 0 };printf("请输入一个数字:\n");scanf("%d", &num);for (i = 0, j = 0; j < 32; ++i, j += 2){arr1[15 - j / 2] = (num >> j) & 1;arr2[15 - i] = (num >> (2 * i + 1)) & 1;}printf("%d对应二进制的偶数位分别是:\n", num);for (i = 0; i < 16; i++){printf("%d", arr1[i]);}printf("\n");printf("%d对应二进制的奇数位分别是:\n", num);for (i = 0; i < 16; i++){printf("%d", arr2[i]);}printf("\n");return 0;
}//3. 输出一个整数的每一位
#include<stdio.h>
int main()
{int i = 0;int num = 0;int arr[32] = { 0 };printf("请输入一个整数:\n");scanf("%d", &num);for (i = 0; i < 32; ++i){arr[31 - i] = (num >> i) & 1;}printf("%d对应的二进制数字是:", num);for (int i = 0; i < 32; ++i){printf("%d", arr[i]);}printf("\n");return 0;
}//4.编程实现：
//两个int（32位）整数m和n的二进制表达中，有多少个位(bit)不同？
//输入例子 :
//1999 2299
//输出例子 : 7
int main()
{int m = 0;int n = 0;int temp = 0;int count = 0;printf("请分别输入两个整数m,n\n");scanf("%d%d", &m, &n);temp = m ^ n;int i = 0;for (i = 0; i < 32; ++i){if ((temp >> i) & 1)++count;}printf("%d和%d二进制表达式中有%d位不同\n", m, n, count);return 0;
}