Detailed introduction to PHP variable reference and object reference_PHP tutorial

The article summarizes how to make variable references in PHP and what are variable references? How to do it? Let’s introduce the usage of PHP variable references one by one.

What does a quote do
PHP's references allow two variables to point to the same content. Meaning, when doing this:

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代码如下	复制代码
$a =& $b; ?>

$a =& $b;

?>

This means $a and $b point to the same variable.

Note:

$a and $b are exactly the same here. It’s not that $a points to $b or vice versa, but that $a and $b point to the same place.

Note:

If an array with a reference is copied, its value will not be dereferenced. The same is true for passing array values ​​to functions.

Note:

If an undefined variable is assigned by reference, passed by reference, or returned by reference, the variable will be automatically created.

Example #1 Using references to undefined variables

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function foo(&$var) { } foo($a); // $a is "created" and assigned to null $b = array(); foo($b['b']); var_dump(array_key_exists('b', $b)); // bool(true) $c = new StdClass; foo($c->d); var_dump(property_exists($c, 'd')); // bool(true) ?>

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function foo(&$var) { }

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$bar =& new fooclass(); $foo =& find_var($bar); ?>

foo($a); // $a is "created" and assigned to null

$b = array();

foo($b['b']);

var_dump(array_key_exists('b', $b)); // bool(true)

$c = new StdClass;

foo($c->d);

var_dump(property_exists($c, 'd')); // bool(true)
?>

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$a="ABC"; $b =&$a; echo $a;//这里输出:ABC echo $b;//这里输出:ABC $b="EFG"; echo $a;//这里$a的值变为EFG 所以输出EFG echo $b;//这里输出EFG ?>

The same syntax can be used in functions, which return references, and in the new operator (PHP 4.0.4 and later):

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$bar =& new fooclass(); $foo =& find_var($bar); ?>

As of PHP 5, new automatically returns a reference, so using =& here is obsolete and produces an E_STRICT level message. Note: Not using the & operator causes a copy of the object to be generated. If you use $this in a class, it will apply to the current instance of that class. Assignment without & will copy the instance (e.g. object) and $this will be applied to the copy, which is not always the desired result. Due to performance and memory consumption issues, you usually only want to work on one instance. Although it is possible to suppress any error messages in a constructor using the @ operator, such as @new, this has no effect when using a &new statement. This is a limitation of the Zend engine and will cause a parsing error. PHP references allow you to use two variables to point to the same content

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$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 ?>

Function call by address
I won’t go into details about call by address. The code is given directly below

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function test(&$a) { $a=$a+100; } $b=1; echo $b;//输出１ test($b); //这里$b传递给函数的其实是$b的变量内容所处的内存地址，通过在函数里改变$a的值　就可以改变$b的值了 echo " "; echo $b;//输出101

function test(&$a)

{

$a=$a+100;

}
$b=1;

echo $b;//output 1

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function &test() { static $b=0;//申明一个静态变量 $b=$b+1; echo $b; return $b; } $a=test();//这条语句会输出　$b的值　为１ $a=5; $a=test();//这条语句会输出　$b的值　为2 $a=&test();//这条语句会输出　$b的值　为3 $a=5; $a=test();//这条语句会输出　$b的值　为6

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

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

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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

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

$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

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 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 $a variable to the same place That is to say, the effect equivalent to this is produced ($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 functions. In fact, the reference return of functions is mostly used in objects Object reference

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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 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 .

The role of quotation
If the program is relatively large, there are many variables referencing 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. 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:

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

$a = 1; $b =& $a;

unset ($a);
?>

Won’t unset $b, just $a.

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

global reference

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

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$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", that is, 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, 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 and automatically generate a data copy of $a for $b and re-apply for a piece of memory for storage

代码如下

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$var1 = "Example variable"; $var2 = ""; function global_references($use_globals) {     global $var1, $var2;     if (!$use_globals) {         $var2 =& $var1; // visible only inside the function     } else {         $GLOBALS["var2"] =& $var1; // visible also in global context     } } global_references(false); echo "var2 is set to '$var2'n"; // var2 is set to '' global_references(true); echo "var2 is set to '$var2'n"; // var2 is set to 'Example variable' ?>

If a reference is assigned to a variable declared global within a function, the reference is visible only within the function. This can be avoided by using the $GLOBALS array. Example #2 Reference global variables within a function

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$var1 = "Example variable"; $var2 = ""; function global_references($use_globals) { global $var1, $var2; If (!$use_globals) { $var2 =& $var1; // visible only inside the function } else { $GLOBALS["var2"] =& $var1; // visible also in global context } } global_references(false); echo "var2 is set to '$var2'n"; // var2 is set to '' global_references(true); echo "var2 is set to '$var2'n"; // var2 is set to 'Example variable' ?>

Think of global $var; as shorthand for $var =& $GLOBALS['var'];. Thus assigning another reference to $var only changes the reference to the local variable.

Let’s look at an interview question below

PHP interview questions are as follows:

Ask:

The code is as follows

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$a = 1; $x =&$a; $b=$a++; ?>

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$a = 1;

$x =&$a;
$b=$a++;
?>

What are the values ​​of $b and $x?

The answers to php interview questions are as follows:

$b = 1; $x = 2; See how much you calculated and how much you know about interview references and object references. http://www.bkjia.com/PHPjc/631279.htmlwww.bkjia.com

true

http: //www.bkjia.com/PHPjc/631279.html

TechArticleThe article summarizes how to make variable references in php and what are variable references? How to do it? Let’s introduce the usage of PHP variable references one by one. What to do with PHP quotes...