A detailed discussion of the uses of bit operations in PHP

The editor below will bring you a commonplace talk about the uses of PHPbit operations. The editor thinks it is quite good, so I will share it with you now and give it as a reference for everyone. Let’s follow the editor and take a look.

In practical applications, user permissions can be used.

The Permission management method I mentioned here is a commonly used method. , mainly using the "bit operator" operation, the & bit and operator , | bit or operator. If the number involved in the operation is a decimal number, it will be converted into a binary number to participate in the operation, and then the calculation result will be converted into a decimal number and output.

Its permission value is like this

2^0=1, the corresponding 2-digit number is "0001" (here^I Expressed as "power", that is: 2 raised to the 0th power, the same below)

2^1=2, the corresponding 2-ary number is "0010"
2^2=4, the corresponding 2-ary number The number is "0100"
2^3=8, and the corresponding binary number is "1000"

To determine whether a number is within a certain number range, you can use the & operator (the value is from the table above Winning)

For example: 7=4|2|1　(You can also simply understand it as 7=4+2+1)

Use & to operate, you can know that 7&4, 7&2, 7&1 are all true, and if 7&8 is false,

&, | If you are not familiar with it, you should check the manual to see how to use it. Okay

Let’s take a look at an example:

// 赋予权限值-->删除：8、上传：4、写入：2、只读：1
define(“mDELETE”,8);
define(“mUPLOAD”,4);
define(“mWRITE”,2);
define(“mREAD”,1);
//vvvvvvvvvvvvv使用说明vvvvvvvvvvvvv
//部门经理的权限为(假设它拥有此部门的所有权限)，| 是位或运行符，不熟悉的就查查资料
echo mDELETE|mUPLOAD|mWRITE|mREAD ,”
“;// 相当于是把上面的权限值加起来：8+4+2+1=15
// 设我只有 upload 和 read 权限，则
echo mUPLOAD|mREAD ,”
“;//相当于是把上传、只读的权限值分别相加：4+1=5
/*
*赋予它多个权限就分别取得权限值相加，又比如某位员工拥有除了删除外的权限其余都拥有，那它的权限值是多少?
*应该是：4+2+1＝7
*明白了怎么赋值给权限吧?
*/
//^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
//判断某人的权限可用，设权限值在$key中
/*
*判断权限用&位与符，
*/
$key = 13;//13＝8+4+1
if($key & mDELETE) echo “有删除权限
“; //8
if($key & mUPLOAD) echo “有上传权限
“; //4
$a=$key & mWRITE; echo “有写权限
“.$a; //无此权限
if($key & mREAD) echo “有读权限
“; //1
?>

OK, among them, the permission score An algorithm is like this, which can be said to be simple and efficient. I don’t know if you understand it or not. It doesn’t matter if you don’t, just remember the examples. The premise is to distribute the permission values ​​well, that is, 1, 2, 4, 8, 16... (There is also an order issue here. The higher the permission, the higher the permission value, such as the deletion demonstrated in the above example. permissions). With the permission distribution table, you can determine what permissions are given to a person. You can simply understand that which permissions you want, just add the corresponding permission values.

This method is very easy to use. The disadvantage is that if the permissions are distributed finely, the permission value will become larger and larger. Think about it for yourself, how many powers of 2, if all the permissions are required? Add it all up. But for general permissions this is enough.

The following are some simple application examples

(1) Determine int typevariablea Is it an odd number or an even number

a&1 = 0 Even number

a&1 = 1 Odd number

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

##(3) will int type The k-th bit of variable a is cleared to 0, that is, a=a&~(1<

(4) will int The k-th position of type variable a is 1, that is, a=a|(1<##<>

(5) int type The variable is rotated left k times, that is, a=a<>16-k (assuming sizeof(int)=16)

For two integers x, y, if you use (x+y)/2 to find the average, overflow will occur because x+y may Greater than INT_MAX, but we know 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);

}

boolean power2(int x)

{

return ((x&(x-1))==0)&&(x!=0)；

}

##

void swap(int x , int y)

{

x ^= y;

y ^= x;

x ^= y;

}

(10) Calculate the absolute value

int abs( int x )

{

int y ;

y = x >> 31 ;

return (x^y)-y ; //or: (x+y)^y

}

(11) Modulo operation is converted into bit operation (without overflow)

a % (2^n) 等价于 a & (2^n – 1)

a * (2^n) 等价于 a<< n

a / (2^n) 等价于 a>> n
例: 12/8 == 12>>3

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

　 else x= a;
　　 等价于 x= a ^ b ^ x;

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