Detailed explanation of php reference (&) symbol

$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 EFG is output
echo $b;//EFG is output here
?>

2, function transfer address transfer I won’t go into details about call by address. The code is given directly below.

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 the value of $a in the function, the value of $b can be changed
echo "" ;
echo $b;//Output 101
?>

Note: If you test(1); here, an error will occur.

3, function reference returns

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
?>

Explanation: In this way, $a=test(); actually does not get a function reference return, which is no different from an ordinary function call. As for the reason: PHP regulations PHP stipulates that what is obtained through $a=&test(); is the reference return of the function. 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.).

$a=test() method of calling a function 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 $a variable to the same place. That is to say, the effect equivalent to this is produced ($a=&b;), so changing the value of $a also changes the value of $b, so after executing

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

The above code is the effect of running in PHP5 In PHP5, object copying is achieved through references. 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 the object and hope that changes to the original object will not affect the copy. For this purpose, PHP defines a special method called __clone .

5, the role of quotes If the program is relatively large, there are many variables referencing the same object, and you want to manually clear the object after using it, I personally recommend using the "&" method, and then using $var=null to clear it. Otherwise, use the default of php5 Method. In addition, for transferring large arrays in php5, it is recommended to use the "&" method, after all, it saves memory space.

6, unquote When you unset a reference, you just break the binding between the variable name and the variable's contents. This does not mean that the variable contents are destroyed. For example:

$a = 1;
$b =& $a;
unset ($a);
?>

will not unset $b, Just $a.

7, global reference When you declare a variable with global $var you actually create a reference to the global variable. It's the same as this:

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

This means that, for example, unset $var will not unset a global variable.

$this In an object method, $this is always a reference to the object that calls it.

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 adopts the principle of "copy-on-write", that is, unless a write operation occurs, it points to the same Address variables or objects will not be copied.

1, if you have the following code

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

In fact, at this time, $a and $b both point to the same memory address, rather than $a and $b occupying different memories

2. If you add the following code to the above code

$a="EFG";

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

This is about the usage of the reference symbol & in php. I hope it can help everyone. Articles you may be interested in: Detailed explanation of examples of php reference passing by value Example code quoted by php Detailed explanation of examples quoted by php Explanation about PHP references Example analysis of PHP variable references, function addresses and object references Detailed introduction to php reference passing by value Understand the difference between passing by value and passing by reference in PHP through examples Look at the efficiency issues of php address reference through examples About the problem of changing the variable value of the php reference address Quotes in PHP, "&" explanation