PHP 5.6.0beta1 released

Que font les références ?

Il existe trois principales utilisations des références : l'assignation par référence, le passage par référence, et le retour par référence. Cette section va introduire ces opérations, avec des liens vers plus de précisions.

Assignation par référence

Dans ce premier cas, les références PHP permettent de faire en sorte que deux variables référencent le même contenu. Par exemple :

<?php
$a 
=& $b;
?>
Cette écriture indique que $a et $b pointent sur le même contenu.

Note:

$a et $b sont complètement égales ici : ce n'est pas $a qui pointe sur $b, ou vice-versa. Ce sont bien $a et $b qui pointent sur le même contenu.

Note:

Si vous assignez, passez, ou retournez, une variable indéfinie par référence, elle sera créée automatiquement.

Exemple #1 Utilisation des références avec des variables indéfinies

<?php
function foo(&$var) { }

foo($a); // $a est "créée" et assignée à 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)
?>

La même syntaxe peut être utilisée avec les fonctions qui retournent des références, et avec l'opérateur new (depuis PHP 4.0.4 et avant PHP 5.0.0) :

<?php
$foo 
=& find_var($bar);
?>
Depuis PHP 5, new retourne automatiquement une référence, donc l'utilisation de =& dans ce contexte est obsolète, et produit un message de niveau E_DEPRECATED à partir de PHP 5.3, et un message E_STRICT avec les versions antérieures. (Techniquement, la difference est qu'en PHP 5, les objets, comme les ressources en fait, sont des pointeurs vers la donnée réelle. La notion de "référence" ici est donc un peu différente de celle utilisée précédemment (alias). Pour plus d'informations, voyez Les objects et les références.)

Avertissement

Si vous assignez une référence à une variable déclarée comme global dans une fonction, la référence ne sera visible qu'au sein de la fonction. Vous pouvez éviter cela en utilisant le tableau $GLOBALS.

Exemple #2 Référencer des variables globales depuis des fonctions

<?php
$var1 
"Variable Exemple";
$var2 "";

function 
global_references($use_globals)
{
    global 
$var1$var2;
    if (!
$use_globals) {
        
$var2 =& $var1// visible uniquement dans la fonction
    
} else {
        
$GLOBALS["var2"] =& $var1// visible également dans le contexte global
    
}
}

global_references(false);
echo 
"var2 est défini à '$var2'\n"// var2 est défini à ''
global_references(true);
echo 
"var2 est défini à '$var2'\n"// var2 est défini à 'Variable Exemple'
?>
Voyez global $var; comme un raccourci pour $var =& $GLOBALS['var'];. De ce fait, assigner une autre référence à $var modifie uniquement la référence locale de la variable.

Note:

Si vous assignez une valeur à une variable qui a des références dans une structure foreach, les références seront également modifiées.

Exemple #3 Références et structure foreach

<?php
$ref 
0;
$row =& $ref;
foreach (array(
123) as $row) {
    
// faites quelque chose
}
echo 
$ref// 3 - le dernier élément du tableau itéré
?>

Même si ce n'est pas strictement une assignation par référence, les expressions créees avec la structure de langage array() peuvent aussi se comporter comme telles, en préfixant par & l'élément du tableau. Voici un exemple :

<?php
$a 
1;
$b = array(23);
$arr = array(&$a, &$b[0], &$b[1]);
$arr[0]++; $arr[1]++; $arr[2]++;
/* $a == 2, $b == array(3, 4); */
?>

Notez par contre que les références à l'intérieur des tableaux peuvent s'avérer dangereuses. Utiliser une assignation normale (pas par référence) avec une référence à droite de l'opérateur ne transforme pas la partie gauche de l'assignation en référence, mais les références à l'intérieur des tableaux sont préservées. Ceci s'applique également aux appels de fonctions avec un tableau passé par valeur. Exemple :

<?php
/* Assignation de variables scalaires */
$a 1;
$b =& $a;
$c $b;
$c 7// $c n'est pas une référence ; pas de changement à $a ou $b

/* Assignation de variables de type tableau */
$arr = array(1);
$a =& $arr[0]; // $a et $arr[0] sont des références vers la même valeur
$arr2 $arr// PAS une assignation par référence!
$arr2[0]++;
/* $a == 2, $arr == array(2) */
/* Les contenus de $arr sont changés même si ce n'était pas une référence ! */
?>
Autrement dit, d'un point de vue des références, le comportement des tableaux est défini élément par élément ; le comportement de chaque élément est indépendant du statut de référence du tableau qui les contient.

Passage par référence

Le deuxième intérêt des références est de permettre de passer des variables par référence. On réalise ceci en faisant référencer le même contenu par une variable locale à un fonction et par une variable du contexte appelant. Par exemple :

<?php
function foo(&$var) {
  
$var++;
}
$a=5;
foo($a);
?>
Après l'exécution de cette portion de code, $a vaut 6. Cela provient du fait que, dans la fonction foo, la variable $var pointe sur le même contenu que $a. Pour plus d'informations à ce sujet, vous pouvez consulter la section passage par référence.

Retour par référence

Le troisième intérêt des références est de permettre le retour de valeurs par référence.

add a note add a note

User Contributed Notes 21 notes

up
8
elrah [] polyptych [dot] com
3 years ago
It appears that references can have side-effects.  Below are two examples.  Both are simply copying one array to another.  In the second example, a reference is made to a value in the first array before the copy.  In the first example the value at index 0 points to two separate memory locations. In the second example, the value at index 0 points to the same memory location.

I won't say this is a bug, because I don't know what the designed behavior of PHP is, but I don't think ANY developers would expect this behavior, so look out.

An example of where this could cause problems is if you do an array copy in a script and expect on type of behavior, but then later add a reference to a value in the array earlier in the script, and then find that the array copy behavior has unexpectedly changed.

<?php
// Example one
$arr1 = array(1);
echo
"\nbefore:\n";
echo
"\$arr1[0] == {$arr1[0]}\n";
$arr2 = $arr1;
$arr2[0]++;
echo
"\nafter:\n";
echo
"\$arr1[0] == {$arr1[0]}\n";
echo
"\$arr2[0] == {$arr2[0]}\n";

// Example two
$arr3 = array(1);
$a =& $arr3[0];
echo
"\nbefore:\n";
echo
"\$a == $a\n";
echo
"\$arr3[0] == {$arr3[0]}\n";
$arr4 = $arr3;
$arr4[0]++;
echo
"\nafter:\n";
echo
"\$a == $a\n";
echo
"\$arr3[0] == {$arr3[0]}\n";
echo
"\$arr4[0] == {$arr4[0]}\n";
?>
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2
Drewseph
5 years ago
If you set a variable before passing it to a function that takes a variable as a reference, it is much harder (if not impossible) to edit the variable within the function.

Example:
<?php
function foo(&$bar) {
   
$bar = "hello\n";
}

foo($unset);
echo(
$unset);
foo($set = "set\n");
echo(
$set);

?>

Output:
hello
set

It baffles me, but there you have it.
up
2
charles at org oo dot com
6 years ago
points to post below me.
When you're doing the references with loops, you need to unset($var).

for example
<?php
foreach($var as &$value)
{
...
}
unset(
$value);
?>
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2
amp at gmx dot info
6 years ago
Something that might not be obvious on the first look:
If you want to cycle through an array with references, you must not use a simple value assigning foreach control structure. You have to use an extended key-value assigning foreach or a for control structure.

A simple value assigning foreach control structure produces a copy of an object or value. The following code

$v1=0;
$arrV=array(&$v1,&$v1);
foreach ($arrV as $v)
{
  $v1++;
  echo $v."\n";
}

yields

0
1

which means $v in foreach is not a reference to $v1 but a copy of the object the actual element in the array was referencing to.

The codes

$v1=0;
$arrV=array(&$v1,&$v1);
foreach ($arrV as $k=>$v)
{
    $v1++;
    echo $arrV[$k]."\n";
}

and

$v1=0;
$arrV=array(&$v1,&$v1);
$c=count($arrV);
for ($i=0; $i<$c;$i++)
{
    $v1++;
    echo $arrV[$i]."\n";
}

both yield

1
2

and therefor cycle through the original objects (both $v1), which is, in terms of our aim, what we have been looking for.

(tested with php 4.1.3)
up
2
Hlavac
6 years ago
Watch out for this:

foreach ($somearray as &$i) {
  // update some $i...
}
...
foreach ($somearray as $i) {
  // last element of $somearray is mysteriously overwritten!
}

Problem is $i contians reference to last element of $somearray after the first foreach, and the second foreach happily assigns to it!
up
2
ladoo at gmx dot at
9 years ago
I ran into something when using an expanded version of the example of pbaltz at NO_SPAM dot cs dot NO_SPAM dot wisc dot edu below.
This could be somewhat confusing although it is perfectly clear if you have read the manual carfully. It makes the fact that references always point to the content of a variable perfectly clear (at least to me).

<?php
$a
= 1;
$c = 2;
$b =& $a; // $b points to 1
$a =& $c; // $a points now to 2, but $b still to 1;
echo $a, " ", $b;
// Output: 2 1
?>
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1
php at hood dot id dot au
7 years ago
I discovered something today using references in a foreach

<?php
$a1
= array('a'=>'a');
$a2 = array('a'=>'b');

foreach (
$a1 as $k=>&$v)
$v = 'x';

echo
$a1['a']; // will echo x

foreach ($a2 as $k=>$v)
{}

echo
$a1['a']; // will echo b (!)
?>

After reading the manual this looks like it is meant to happen. But it confused me for a few days!

(The solution I used was to turn the second foreach into a reference too)
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1
Amaroq
4 years ago
I think a correction to my last post is in order.

When there is a constructor, the strange behavior mentioned in my last post doesn't occur. My guess is that php was treating reftest() as a constructor (maybe because it was the first function?) and running it upon instantiation.

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
__construct()
    {
        return
0;
    }

    public function
reftest()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 2.
echo $reference->c; //Echoes 2.
?>
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0
Amaroq
4 years ago
When using references in a class, you can reference $this-> variables.

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest()
    {
       
$b =& $this->a;
       
$b = 2;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 2.
echo $reference->c; //Echoes 2.
?>

However, this doesn't appear to be completely trustworthy. In some cases, it can act strangely.

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 3.
echo $reference->c; //Echoes 2.
?>

In this second code block, I've changed reftest() so that $b increments instead of just gets changed to 2. Somehow, it winds up equaling 3 instead of 2 as it should.
up
0
php.devel at homelinkcs dot com
9 years ago
In reply to lars at riisgaardribe dot dk,

When a variable is copied, a reference is used internally until the copy is modified.  Therefore you shouldn't use references at all in your situation as it doesn't save any memory usage and increases the chance of logic bugs, as you discoved.
up
-1
admin at torntech dot com
7 months ago
Something that has not been discussed so far is reference of a reference.
I needed a quick and dirty method of aliasing incorrect naming until a proper rewrite could be done.
Hope this saves someone else the time of testing since it was not covered in the Does/Are/Are Not pages.
Far from best practice, but it worked.

<?php
$a
= 0;

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

$a = 5;
echo
$a . ', ' . $b;
//ouputs: 5,5

echo ' | ';

$b = 6;
echo
$a . ',' . $b;
//outputs: 6,6

echo ' | ';
unset(
$a );
echo
$a . ', ' . $b;

//outputs: , 6

class Product {

    public
$id;
    private
$productid;

    public function
__construct( $id = null ) {
       
$this->id =& $this->productid;
       
$this->productid =& $this->id;
       
$this->id = $id;
    }

    public function
getProductId() {
        return
$this->productid;
    }

}

echo
' | ';

$Product = new Product( 1 );
echo
$Product->id . ', ' . $Product->getProductId();
//outputs 1, 1
$Product->id = 2;
echo
' | ';
echo
$Product->id . ', ' . $Product->getProductId();
//outputs 2, 2
$Product->id = null;
echo
' | ';
echo
$Product->id . ', ' . $Product->getProductId();
//outouts ,
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-2
butshuti at smartrwanda dot org
1 year ago
This appears to be the hidden behavior: When a class function has the same name as the class, it seems to be implicitly called when an object of the class is created.
For instance, you may take a look at the naming of the function "reftest()": it is in the class "reftest". The behavior can be tested as follows:

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest1()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
   
    public function
reftest()
    {
       echo
"REFTEST() called here!\n";
    }
}

$reference = new reftest();
/*You must notice the above will also implicitly call reference->reftest()*/

$reference->reftest1();
$reference->reftest2();

echo
$reference->a."\n"; //Echoes 2, not 3 as previously noticed.
echo $reference->c."\n"; //Echoes 2.
?>

The above outputs:

REFTEST() called here!
2
2

Notice that reftest() appears to be called (though no explicit call to it was made)!
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-2
nay at woodcraftsrus dot com
2 years ago
in PHP you don't really need pointer anymore if you want to share an  object across your program

<?php
class foo{
        protected
$name;
        function
__construct($str){
               
$this->name = $str;
        }
        function
__toString(){
                return 
'my name is "'. $this->name .'" and I live in "' . __CLASS__ . '".' . "\n";
        }
        function
setName($str){
               
$this->name = $str;
        }
}

class
MasterOne{
        protected
$foo;
        function
__construct($f){
               
$this->foo = $f;
        }
        function
__toString(){
                return
'Master: ' . __CLASS__ . ' | foo: ' . $this->foo . "\n";
        }
        function
setFooName($str){
               
$this->foo->setName( $str );
        }
}

class
MasterTwo{
        protected
$foo;
        function
__construct($f){
               
$this->foo = $f;
        }
        function
__toString(){
                return
'Master: ' . __CLASS__ . ' | foo: ' . $this->foo . "\n";
        }
        function
setFooName($str){
               
$this->foo->setName( $str );
        }
}

$bar = new foo('bar');

print(
"\n");
print(
"Only Created \$bar and printing \$bar\n");
print(
$bar );

print(
"\n");
print(
"Now \$baz is referenced to \$bar and printing \$bar and \$baz\n");
$baz =& $bar;
print(
$bar );

print(
"\n");
print(
"Now Creating MasterOne and Two and passing \$bar to both constructors\n");
$m1 = new MasterOne( $bar );
$m2 = new MasterTwo( $bar );
print(
$m1 );
print(
$m2 );

print(
"\n");
print(
"Now changing value of \$bar and printing \$bar and \$baz\n");
$bar->setName('baz');
print(
$bar );
print(
$baz );

print(
"\n");
print(
"Now printing again MasterOne and Two\n");
print(
$m1 );
print(
$m2 );

print(
"\n");
print(
"Now changing MasterTwo's foo name and printing again MasterOne and Two\n");
$m2->setFooName( 'MasterTwo\'s Foo' );
print(
$m1 );
print(
$m2 );

print(
"Also printing \$bar and \$baz\n");
print(
$bar );
print(
$baz );
?>
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-2
akinaslan at gmail dot com
3 years ago
In this example class name is different from its first function and however there is no construction function. In the end as you guess "a" and "c" are equal. So if there is no construction function at same time class and its first function names are the same, "a" and "c" doesn't equal forever. In my opinion php doesn't seek any function for the construction as long as their names differ from each others.

<?php
class reftest_new
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest_new();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 2.
echo $reference->c; //Echoes 2.
?>
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-2
Amaroq
6 years ago
The order in which you reference your variables matters.

<?php
$a1
= "One";
$a2 = "Two";
$b1 = "Three";
$b2 = "Four";

$b1 =& $a1;
$a2 =& $b2;

echo
$a1; //Echoes "One"
echo $b1; //Echoes "One"

echo $a2; //Echoes "Four"
echo $b2; //Echoes "Four"
?>
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-2
joachim at lous dot org
11 years ago
So to make a by-reference setter function, you need to specify reference semantics _both_ in the parameter list _and_ the assignment, like this:

class foo{
   var $bar;
   function setBar(&$newBar){
      $this->bar =& newBar;
   }
}

Forget any of the two '&'s, and $foo->bar will end up being a copy after the call to setBar.
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-2
dovbysh at gmail dot com
6 years ago
Solution to post "php at hood dot id dot au 04-Mar-2007 10:56":

<?php
$a1
= array('a'=>'a');
$a2 = array('a'=>'b');

foreach (
$a1 as $k=>&$v)
$v = 'x';

echo
$a1['a']; // will echo x

unset($GLOBALS['v']);

foreach (
$a2 as $k=>$v)
{}

echo
$a1['a']; // will echo x

?>
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-3
dnhuff at acm dot org
5 years ago
In reply to Drewseph using foo($a = 'set'); where $a is a reference formal parameter.

$a = 'set' is an expression. Expressions cannot be passed by reference, don't you just hate that, I do. If you turn on error reporting for E_NOTICE, you will be told about it.

Resolution: $a = 'set'; foo($a); this does what you want.
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-3
firespade at gmail dot com
7 years ago
Here's a good little example of referencing. It was the best way for me to understand, hopefully it can help others.

$b = 2;
$a =& $b;
$c = $a;
echo $c;

// Then... $c = 2
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-2
Oddant
10 months ago
About the example on array references.
I think this should be written in the array chapter as well.
Indeed if you are new to programming language in some way, you should beware that arrays are pointers to a vector of Byte(s).

<?php $arr = array(1); ?>
$arr here contains a reference to which the array is located.
Writing :
<?php echo $arr[0]; ?>
dereferences the array to access its very first element.

Now something that you should also be aware of  (even you are not new to programming languages) is that PHP use references to contains the different values of an array. And that makes sense because the type of the elements of a PHP array can be different.

Consider the following example :

<?php

$arr
= array(1, 'test');

$point_to_test =& $arr[1];

$new_ref = 'new';

$arr[1] =& $new_ref;

echo
$arr[1]; // echo 'new';
echo $point_to_test; // echo 'test' ! (still pointed somewhere in the memory)

?>
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-5
strata_ranger at hotmail dot com
4 years ago
An interesting if offbeat use for references:  Creating an array with an arbitrary number of dimensions.

For example, a function that takes the result set from a database and produces a multidimensional array keyed according to one (or more) columns, which might be useful if you want your result set to be accessible in a hierarchial manner, or even if you just want your results keyed by the values of each row's primary/unique key fields.

<?php
function array_key_by($data, $keys, $dupl = false)
/*
 * $data  - Multidimensional array to be keyed
 * $keys  - List containing the index/key(s) to use.
 * $dupl  - How to handle rows containing the same values.  TRUE stores it as an Array, FALSE overwrites the previous row.
 *         
 * Returns a multidimensional array indexed by $keys, or NULL if error.
 * The number of dimensions is equal to the number of $keys provided (+1 if $dupl=TRUE).
 */  
{
   
// Sanity check
   
if (!is_array($data)) return null;
   
   
// Allow passing single key as a scalar
   
if (is_string($keys) or is_integer($keys)) $keys = Array($keys);
    elseif (!
is_array($keys)) return null;

   
// Our output array
   
$out = Array();
   
   
// Loop through each row of our input $data
   
foreach($data as $cx => $row) if (is_array($row))
    {
     
     
// Loop through our $keys
     
foreach($keys as $key)
      {
       
$value = $row[$key];

        if (!isset(
$last)) // First $key only
       
{
          if (!isset(
$out[$value])) $out[$value] = Array();
         
$last =& $out; // Bind $last to $out
       
}
        else
// Second and subsequent $key....
       
{
          if (!isset(
$last[$value])) $last[$value] = Array();
        }

       
// Bind $last to one dimension 'deeper'.
        // First lap: was &$out, now &$out[...]
        // Second lap: was &$out[...], now &$out[...][...]
        // Third lap:  was &$out[...][...], now &$out[...][...][...]
        // (etc.)
       
$last =& $last[$value];
      }
     
      if (isset(
$last))
      {
       
// At this point, copy the $row into our output array
       
if ($dupl) $last[$cx] = $row; // Keep previous
       
else       $last = $row; // Overwrite previous
     
}
      unset(
$last); // Break the reference
   
}
    else return
NULL;
   
   
// Done
   
return $out;
}

// A sample result set to test the function with
$data = Array(Array('name' => 'row 1', 'foo' => 'foo_a', 'bar' => 'bar_a', 'baz' => 'baz_a'),
              Array(
'name' => 'row 2', 'foo' => 'foo_a', 'bar' => 'bar_a', 'baz' => 'baz_b'),
              Array(
'name' => 'row 3', 'foo' => 'foo_a', 'bar' => 'bar_b', 'baz' => 'baz_c'),
              Array(
'name' => 'row 4', 'foo' => 'foo_b', 'bar' => 'bar_c', 'baz' => 'baz_d')
              );

// First, let's key it by one column (result: two-dimensional array)
print_r(array_key_by($data, 'baz'));

// Or, key it by two columns (result: 3-dimensional array)
print_r(array_key_by($data, Array('baz', 'bar')));

// We could also key it by three columns (result: 4-dimensional array)
print_r(array_key_by($data, Array('baz', 'bar', 'foo')));

?>
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