Binary
Binary 1 and 0 is used to refer to a computer language, 0 may represent any number, which is closed on a binary
The decimal (i.e. digits) is converted into a binary calculation step
There are three key concepts in binary
▶ bit computing
Bitwise operation rules:
Bitwise AND &: Two all 1, the result is 1
Bitwise or |: there is a two to one, the result is 1
Bitwise XOR ^: two in one of 0, a is 1, the result is a
Bitwise ~: take 0 0 Take 1,1
According to the rules, calculating a number of steps:
Calculation example: 2 =?
1. Locate 2's complement, [because of the way the computer is operational complement to carry out operations]
2 of the original code: 00000000 00000000 0,000,000,000,000,010
// string from left to right above the first number is the number 0, which is a positive number, and a positive there is a principle: positive original code symbol = = trans complement, the complement is above 2 this series of numbers
2. And then negated
2 ~ negated: 1,111,111,111,111,111 1,111,111,111,111,101
// get a string above complement, in order to get the results we want will complement -> anti-code -> original code
3. Complement -> inverted [(complement -1) is inverted]
Anti-code: 1,111,111,111,111,111 1,111,111,111,111,100
4. inverted -> original code [symbol bit change, i.e. from left to right above the first string constant number]
Original code: 1,000,000,000,000,000 0,000,000,000,000,011 = 0 + 1 * 1 * 2 ^ 2 ^ 2 = 1 + 1 = -3 // 1 for a negative front of the red, the result is -3
//^表示次方，2^0表示2的0次方，从右往左开始计算：即1(上面这串数字从右往左数的第一个1)*2^0+1(上面这串数字从右往左数的第二个1)*2^1
例如：2&3=？
1.找到2和3的补码
2的补码： 00000000 00000000 00000000 00000010
3的补码： 00000000 00000000 00000000 00000011
//按位与&的运算：两位都为1，结果为1
2&3= 00000000 00000000 00000000 00000010 // 前面的字串为补码，由于其第一个数字为0，所以其为正数,正数的补码=原码，所以结果为2
终上：计算一个数的步骤就是原码-->反码-->补码-->根据位运算符计算得到补码-->反码-->原码-->再得到我们想要的值
▶ 位移运算
在php中位移运算符有两种：>>(右移)和<<(左移)
运算的规则：
右移：低位溢出，符号位不变，并用符号位补溢出的高位 [通俗点就是将最右边的数溢出，用最左边的数（符号数）补溢出的个数，放在最左边]
左移：符号位不变，低位补0 [通俗点就是将最左边的数溢出，用0来补溢出的个数，放在最右边]
根据规则,下面写几个案例：
A. $a=1>>2;//将1向右移动两位
1.找出1的补码
00000000 00000000 00000000 00000001
1>>2
00000000 00000000 00000000 00000000
$a=1>>2=0
B. $a=1<<2;//将1向左移动两位
1.找出1的补码
00000000 00000000 00000000 00000001
1<<2
00000000 00000000 00000000 00000100
$a=1<<2=4