There are three basic operations performed using references: assigning by reference, passing by reference, and returning by reference. This section will give an introduction to these operations, with links for further reading.
In the first of these, PHP references allow you to make two variables refer to the same content. Meaning, when you do:
<?php
$a =& $b;
?>
Notă:
$a and $b are completely equal here. $a is not pointing to $b or vice versa. $a and $b are pointing to the same place.
Notă:
If you assign, pass, or return an undefined variable by reference, it will get created.
Example #1 Using references with undefined variables
<?php
function foo(&$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)
?>
The same syntax can be used with functions that return references:
<?php
$foo =& find_var($bar);
?>
E_DEPRECATED message in PHP 5.3 and
later, and an E_STRICT message in earlier versions.
As of PHP 7.0 it is syntactically invalid.
(Technically, the difference is that, in PHP 5, object variables, much like
resources, are a mere pointer to the actual object data, so these object
references are not "references" in the same sense used before (aliases).
For more information, see Objects
and references.)
If you assign a reference to a variable declared global inside a function, the reference will be visible only inside the function. You can avoid this by using the $GLOBALS array.
Example #2 Referencing global variables inside functions
<?php
$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'
?>
Notă:
If you assign a value to a variable with references in a foreach statement, the references are modified too.
Example #3 References and foreach statement
<?php
$ref = 0;
$row =& $ref;
foreach (array(1, 2, 3) as $row) {
// do something
}
echo $ref; // 3 - last element of the iterated array
?>
While not being strictly an assignment by reference, expressions created with the language construct array() can also behave as such by prefixing & to the array element to add. Example:
<?php
$a = 1;
$b = array(2, 3);
$arr = array(&$a, &$b[0], &$b[1]);
$arr[0]++; $arr[1]++; $arr[2]++;
/* $a == 2, $b == array(3, 4); */
?>
Note, however, that references inside arrays are potentially dangerous. Doing a normal (not by reference) assignment with a reference on the right side does not turn the left side into a reference, but references inside arrays are preserved in these normal assignments. This also applies to function calls where the array is passed by value. Example:
<?php
/* Assignment of scalar variables */
$a = 1;
$b =& $a;
$c = $b;
$c = 7; //$c is not a reference; no change to $a or $b
/* Assignment of array variables */
$arr = array(1);
$a =& $arr[0]; //$a and $arr[0] are in the same reference set
$arr2 = $arr; //not an assignment-by-reference!
$arr2[0]++;
/* $a == 2, $arr == array(2) */
/* The contents of $arr are changed even though it's not a reference! */
?>
The second thing references do is to pass variables by reference. This is done by making a local variable in a function and a variable in the calling scope referencing the same content. Example:
<?php
function foo(&$var)
{
$var++;
}
$a=5;
foo($a);
?>
The third thing references can do is return by reference.