XOR operation:
First of all, XOR means that when the binary representation of two numbers is performed XOR operation, if the two binary representations of the current bit are different, it is 1 and the same is 0. This method is widely used to count the number of digits of 1 in a number!
The two values involved in the operation, if the two corresponding bits are the same, the result is 0, otherwise it is 1.
That is:
0^0 = 0,
1^0 = 1,
0^1 = 1,
1^1 = 0
3 characteristics of bitwise XOR:
(1) 0^0=0,0^1=1 0 XOR or any number = any number
(2) 1^0=1,1^1=0 1 XOR any number - negate any number
(3) any number XOR itself = set itself to 0
bitwise XOR several Common uses:
(1) To flip some specific bits
such as the 2nd and 3rd bits of the logarithm 10100001, then the number can be XORed with 00000110.
10100001^00000110 = 10100111
(2) To achieve the exchange of two values without using temporary variables.
For example, exchanging the values of two integers a=10100001 and b=00000110 can be achieved by the following statements:
a = a^b; //a=10100111
b = b^a; //b=10100001
a = a^b; / /a=00000110
bit operation
In bit operation, after the number is represented in binary, the operation of 0 or 1 on each bit is performed. The first step to understanding bit operations is to understand binary. Binary means that each digit of the number is 0 or 1. For example, 2 in decimal is 10 after being converted into binary.
In fact, binary operations are not difficult to master, because there are only 5 kinds of bit operations: AND, OR, XOR, left shift, right shift. The following table:
| and(&) | 0 & 0 = 0 | 1 & 0 = 0 | 0 & 1 = 0 | 1 & 1 = 1 |
| or (|) | 0 | 0 = 0 | 1 | 0 = 1 | 0 | 1 = 1 | 1 | 1 = 1 |
| XOR (^) | 0 ^ 0 = 0 | 1 ^ 0 = 1 | 0 ^ 1 = 1 | 1 ^ 1 = 0 |
Left shift operation :
The left shift operator m<<n means to shift m to the left by n bits. When shifting n bits to the left, the leftmost n bits will be discarded, and n 0s will be added to the rightmost. For example:
00001010 << 2 = 00101000 10001010 << 3 = 01010000
Right shift operation :
The right shift operator m>>n means to shift m to the right by n bits. When shifting right by n bits, the rightmost n bits are discarded. But handling the leftmost bit when shifting right is a little more complicated. Note here that if the number is an unsigned value , the leftmost n bits are padded with 0. If the number is a signed numeric value , the leftmost n bits are padded with the sign bit of the number. That is to say, if the number was originally a positive number, then shift to the right and then add n 0s to the left; if the number was originally negative, then shift to the left and then add n 1s to the left. The following is a heap of two 8-bit signed Example of right-shifting a number:
00001010 >> 2 = 00000010 10001010 >> 3 = 11110001
There is an equivalence relation about shifting operations: shifting an integer to the right by one bit is mathematically equivalent to dividing an integer by 2.
a << = 1 ; // Shifting a to the left is equivalent to a = a * 2; a <<= 2 ; // Shifting a left by 2 bits is equivalent to a = a * 2 to the power of 2 (4);
计算机内部只识别1、0，十进制需变成二进制才能使用移位运算符<<,>> 。
int j = 8; p = j << 1; cout<<p<<endl;
在这里，8左移一位就是8*2的结果16 。
移位运算是最有效的计算乘/除乘法的运算之一。
按位与（&）其功能是参与运算的两数各对应的二进制位相与。只有对应的两个二进制位均为1时，结果位才为1，否则为0 。参与运算的数以补码方式出现。
先举一个例子如下：
题目：请实现一个函数，输入一个正数，输出该数二进制表示中1的个数。
这里用到了这样一个知识点：把一个整数减去1，再和原整数做与运算，会把该整数最右边一个1变成0 。 那么一个整数的二进制表示中有多少个1，就可以进行多少次这样的操作。
总结：把一个整数减去1之后再和原来的整数做位与运算，得到的结果相当于是把整数的二进制表示中的最右边一个1变成0 。
位运算的应用可以运用于很多场合：
- 清零特定位（mask中特定位置0，其它位为1 ， s = s & mask）。
- 取某数中指定位（mask中特定位置，其它位为0, s = s & mask）。
举例：输入两个整数m和n，计算需要改变m的二进制表示中的多少位才能得到n。
解决方法：第一步，求这两个数的异或；第二步，统计异或结果中1的位数。
接下来我们再举一例，就可以更好的说明移位运算了：用一条语句判断一个整数是不是2的整数次方。
解决方法：一个整数如果是2的整数次方，那么它的二进制表示中有且只有一位是1，而其它所有位都是0 。 根据前面的分析，把这个整数减去1后再和它自己做与运算，这个整数中唯一的1就变成0了。
解答：!（x & (x - 1)）