PHP variables, functions, and object references_PHP tutorial-PHP Tutorial-php.cn

PHP variables, functions, and object references_PHP tutorial

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Release： 2016-07-21 14:51:59

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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 references allow you to use two variables to point to the same content
[php]
$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 ?>
[/php]

Function call by address
I won’t go into details about call by address. The code is given directly below
[php]
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
[/php]
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
[php]
function &test()
{
static $b=0;//Declare a static variable
$b=$b+1;
echo $b;
return $b;
}

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

$a=&test();//This statement will output that the value of $b is 3
$a=5;
$a=test();//This statement will output that the value of $b is 6
[/php]
Explain 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. As for 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 not understand it for a long time

To explain using the above example,
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 $a variable to the same place
That is to say, the effect equivalent to this is produced ($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 the function. In fact, the reference return of the function is mostly used in objects

Object reference
[php]
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 ?>
[/php]
The above code is the running effect 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 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.

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);
?>

Doesn'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’s terms
1: If there is the following code
[php]
$a="ABC";
$b=$a;
[/php]
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
[php]
$a="EFG";
[/php]
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