1，/ 与 %的应用
int main()
{
int a = 5 / 2;//商2余1
printf("a = %d\n", a);
int b = 5 % 2;
printf("b = %d\n", b);
return 0;
}
2，移位操作符
右移操作符
1.算术右移
右边丢弃，左边补原符号位
2.逻辑右移
右边丢弃，左边补0
左移操作符
左边丢弃，右边补0
int main()
{
int a = 5;
a << 1;
int b = a << 1;
//00000000000000000000000000101
printf("%d\n", b);
return 0;
}
警告：对于移位运算符，不要移动负数位，这个标准是未定义的。例如：
int num = 10;
num>>-1//error
3，位操作符
int main()
{
//& - 按2进制位与
int a = 3;
int b = 5;
int c = a&b;
//00000000000000000000000000011
//00000000000000000000000000101
//00000000000000000000000000001
printf("%d\n", c);
return 0;
}
//| - 按2进制位或
int a = 3;
int b = 5;
int c = a | b;
////00000000000000000000000000011
////00000000000000000000000000101
////00000000000000000000000000111
printf("%d\n", c);
return 0;
}
int a = 3;
int b = 5;
int c = a ^ b;
printf("%d\n", c);
//00000000000000000000000000011
//00000000000000000000000000101
//00000000000000000000000000110
return 0;
int main()
{
int a = 3;
int b = 5;
//int tmp = 0;//临时变量
printf("before: a = %d b = %d\n", a, b);
//tmp = a;
//a = b;
//b = tmp;
//加减法 缺陷：可能会溢出
/*a = a + b;
b = a - b;
a = a - b;*/
a = a^b;
b = a^b;
a = a^b;
printf("after: a = %d b =%d\n", a, b);
return 0;
}
练习1
int main()
{
int num = 0;
int count = 0;
scanf_s("%d", &num);
int i = 0;
for (i = 0; i < 32; i++)
{
if (1 == ((num >> i) & 1))
count++;
}
printf("%d\n", count);
//统计num的补码中有几1
//while (num)
//{
// if (num % 2 == 1)
// count++;
// num = num / 2;
//}
return 0;
}
4，赋值操作符 =
复合赋值符
int main()
{
int a = 10;
a = a + 2;
a += 2;//复合赋值符
a = a >> 1;
a >>= 1;
a = a & 1;
a &= 1;
return 0;
}
按位取反~
int main()
{
int a = 11;
a = a | (1 << 2);
printf("%d\n", a);
a = a & (~(1 << 2));
printf("%d\n", a);
return 0;
}
不写了哥们我累死了
