PHP reference finishing (2.29)_PHP tutorial

Quotation in PHP means: different names access the same variable content.
It is different from pointers in C language. The pointer in C language stores the address where the content of the variable is stored in memory

Variable reference

PHP’s references allow you to use two variables to point to the same content

$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;//Output EFG
here or

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

here

?>

Function call by address
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. The value of $b can be changed by changing the value of $a in the function
echo "
";
echo $b;//Output 101

It should be noted that if test(1); is used here, an error will occur. You have to think about the reason yourself

Function reference returns
Let’s look at the code first

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 below:
In this way, $a=test(); actually does not get a reference return from the function. It is no different from an ordinary function call. The reason: This is the regulation of PHP
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.) This nonsense made me unable to understand it for a long time www.2cto.com

Using the above example to explain it is
Calling a function using $a=test() 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 $a variable to the same place
That is to say, the effect is equivalent to this ($a=&b;), so changing the value of $a also changes the value of $b, so after executing
$a=&test();
$a=5;
From now on, the value of $b becomes 5

Static variables are used here to let everyone understand the reference return of functions. In fact, the reference return of functions is mostly used in objects

Object reference

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 effect of running in PHP5
In PHP5, object copying is achieved by reference. 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 .

The role of quotation
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.

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);
?>

Won’t unset $b, just $a.

global quote
When you declare a variable with global $var you actually create a reference to the global variable. That is the same as doing 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.

//Another little episode below
The address pointing (similar to a pointer) function 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", which means that unless a write operation occurs, it points to the same address. Variables or objects will not be copied.

In layman terms
1: If there is the following code

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

In fact, $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

$a="EFG";

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

Excerpted from chaojie2009’s column

http://www.bkjia.com/PHPjc/478325.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/478325.htmlTechArticleThe meaning of reference in PHP is: different names access the same variable content. The same as the pointer in C language is There is a difference. Pointers in C language store the contents of variables in memory...