What References Do

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.

Assign By Reference

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;

?>

it means that $a and $b point to the same content.

Зауваження:
$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.

Зауваження:
If you assign, pass, or return an undefined variable by reference, it will get created.

Приклад #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);

?>

Using the same syntax with a function that does not return by reference will give an error, as will using it with the result of the new operator. Although objects are passed around as pointers, these are not the same as references, as explained under Objects and references.

Увага

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.

Приклад #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'

?>

Think about global $var; as a shortcut to

$var
        =& $GLOBALS['var'];

. Thus assigning another reference to $var only changes the local variable's reference.

Зауваження:
If you assign a value to a variable with references in a foreach statement, the references are modified too.

Приклад #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! */

?>

In other words, the reference behavior of arrays is defined in an element-by-element basis; the reference behavior of individual elements is dissociated from the reference status of the array container.

Pass By 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);

?>

will make $a to be 6. This happens because in the function foo the variable $var refers to the same content as $a. For more information on this, read the passing by reference section.

Return By Reference

The third thing references can do is return by reference.