Detailed explanation of the use of reference symbol (&) in php

Detailed explanation of the use of reference symbol (&) in php_PHP tutorial

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Release： 2016-07-13 10:25:10

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It is different from pointers in C language. The pointer in the C language stores a reference to the address variable where the content of the variable is stored in memory.
PHP’s reference allows you to use two variables to point to the same content

Copy code The code is as follows:

$a="ABC";
$b =&$a;
echo $a;//Output here: ABC
echo $b;//Output here: ABC
$b="EFG";
echo $a;//The value of $a here becomes EFG, so the output is EFG
echo $b ;//Output EFG here

Pass-by-reference call of the function
I won’t go into details about the pass-by-reference call. The code is given directly below

Copy Code The code is as follows:

function test(&$a)
{
$a=$a+100;
}
$b =1;
echo $b;//Output 1
test($b); //What $b is passed to the function here is actually the memory address where the variable content of $b is located. By changing it in the function The value of $a can change the value of $b
echo "
";
echo $b;//Output 101

It should be noted that here test(1); an error will occur, think about the reason yourself
The reference return of the function
Look at the code first

Copy the code The code is as follows:

function &test()
{
static $b=0;//Declare a static variable
$b=$b+1;
echo $b;
return $b;
}

$a=test();//This statement will output the value of $b as 1
$a=5;
$a =test();//This statement will output the value of $b as 2

$a=&test();//This statement will output the value of $b as 3
$a= 5;
$a=test();//This statement will output the value of $b as 6

The following explanation:
In this way $a=test( ; 🎜>As for what is a reference return (the PHP manual says: Reference return is used when you want to use a function to find which variable the reference should be bound to.) This nonsense made me unable to understand it for a long time
Use The above example explains that calling a function in the
$a=test() method only assigns the value of the function to $a, and any changes to $a will not affect $b in the function
When calling a function through $a=&test(), its function is to point the memory address of the $b variable in return $b and the memory address of the $a variable to the same place
, which is equivalent to this. Effect ($a=&b;) So changing the value of $a also changes the value of $b. So after executing
$a=&test();
$a=5;
, $ The value of b has changed to 5
Static variables are used here to let everyone understand the reference return of the function. In fact, the reference return of the function is mostly used in objects
Reference of the object

Copy code The code is as follows:

class a{

var $abc="ABC";
}
$b=new a;
$c=$b;
echo $b->abc;//Output ABC here
echo $c->abc;//Output ABC here
$b->abc= "DEF";
echo $c->abc;//Output DEF here

The above code is the running effect in PHP5

In PHP5, the copy of the object is through reference to achieve. In the above column, $b=new a; $c=$b; is actually equivalent to $b=new a; $c=&$b;
The default in PHP5 is to call objects by reference, but sometimes you may want to Create a copy of an object and hope that changes to the original object will not affect the copy. For this purpose, PHP defines a special method called __clone.
The role of references
If the program is relatively large, If there are many variables that refer to the same object, and you want to clear it manually after using the object, I personally recommend using the "&" method, and then using $var=null to clear it. In other cases, use the default method of php5. In addition, For the transfer of large arrays in php5, it is recommended to use the "&" method, after all, it saves memory space.
Unreference
When you unset a reference, you just break the binding between the variable name and the variable content. This does not mean that the variable contents are destroyed. For example:

Copy code The code is as follows:

$a = 1;

$b =& $a;
unset ($a);

will not unset $b, just $a.

global reference
When declaring a variable with global $var, a reference to the global variable is actually established. In other words, it is the same as doing this:

Copy the code The code is as follows:

$var =& $GLOBALS[" var"];

This means that, for example, unset $var will not unset a global variable.
$this
In a method of an object, $this is always a reference to the object that calls it.
//Here’s another little episode
The pointing (similar to pointer) function of the address in PHP is not implemented by the user himself, but is implemented by the Zend core. The reference in PHP uses "copy-on-write" The principle is that unless a write operation occurs, variables or objects pointing to the same address will not be copied.
In layman terms
1: If there is the following code

Copy the codeThe code is as follows:

$a ="ABC";
$b=$a;

In fact, both $a and $b point to the same memory address at this time, rather than $a and $b occupying different memories 2 :If you add the following code to the above code

Copy code The code is as follows:

$a="EFG ";

Since the data in the memory pointed to by $a and $b needs to be rewritten, at this time the Zend core will automatically determine and automatically produce a data copy of $a for $b and apply again. A piece of memory for storage