Examples summarizing the usage skills of PHP bit operators

Explain through an example:

<?php
//1.位运算符两边的值必须是整形和浮点型，当是其它类型的值时会先转换成整形和浮点型再来参与位运算；
 //而逻辑运算符两边参与运算的值必须是布尔型；
 var_dump(12&13);//输出 int 12
 //12转换成32位的二进制数为：00000000 00000000 00000000 00001100
 //13转换成32位的二进制数为：00000000 00000000 00000000 00001101
 //按位运算后得到的值为      00000000 00000000 00000000 00001100 ，等于12
 
 var_dump(&#39;A&#39;&&#39;a&#39;);//输出string &#39;A&#39;;因为A=65，a=97
 
 var_dump(&#39;A&#39;&97);//输出int 0;因为字符串A会先转换为整数的0后再参与位运算
 
 //2.位运算与逻辑运算不同没有短路特性
 $a=3;
 $b=10;
 if($a>5&&$b++<100)
 {
     echo "1111111111";
 }
 echo $b."<br>";//输出$b=10,逻辑运算短路，$b没有自加；
 
 if($a>5&$b++<100)
 {
     echo "1111111";
 }
 echo $b."<br>";//输出$b=11,位运算不短路，$b自加；
 
 //3.按位非，按位异或等比较简单，这里不做过多讲叙；现在讲讲按位左移和按位右移动
 var_dump(12>>2);//输出int 3
 var_dump(12<<2);//输出int 48
 //可以发现，左移几位就相当于乘以2的多少次方；按位右移就相当于除以2的多少次方；
 
?>

(1) Determine whether the int type variable a is an odd number or an even number

a&1 = 0 even number

a&1 = 1 odd number

(2) Take the k-th bit of int type variable a (k=0,1,2...sizeof(int)), that is, a>>k&1

(3) Change the int type variable Clear the k-th bit of a to 0, that is, a=a&~(1<

<>

(4) Set the k-th bit of int type variable a to 1, that is, a=a |(1<

<>

(5) The int type variable is circularly shifted to the left k times, that is, a=a<>16-k (assume sizeof(int)=16 )

(6) The int type variable a is shifted to the right k times in a loop, that is, a=a>>k|a<<16-k (assume sizeof(int)=16)

(7) Average of integers

For two integers x, y, if you use (x+y)/2 to find the average, overflow will occur because x+y may be greater than INT_MAX, but we Knowing that their average value will definitely not overflow, we use the following algorithm:

int average(int x, int y) //返回X,Y 的平均值 
{ 
    return (x&y)+((x^y)>>1); 
}

(8) Determine whether an integer is a power of 2. For a number x >= 0, determine whether it is 2 The power of Without overflow)

a % (2^n) is equivalent to a & (2^n – 1)

(12) The multiplication operation is converted into a bit operation (without In the case of overflow)

a * (2^n) is equivalent to a<< n

(13) The division operation is converted into a bit operation (in the case of no overflow )

a / (2^n) is equivalent to a>> n

Example: 12/8 == 12>>3

(14) a % 2 is equivalent to a & 1

(15) if (x == a) x= b;

　 else x= a;

　 Equivalent to x= a ^ b ^ x;

(16) The opposite number of In the case of 32 bits, shift left

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