Php/docs/language.references.whatdo
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.
Note:
$a
and$b
are completely equal here.$a
is not pointing to$b
or vice versa.$a
and$b
are pointing to the same place.
Note:
If you assign, pass, or return an undefined variable by reference, it will get created.
Example #1 Using references with undefined variables
<?phpfunction 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);?>
new returns a reference automatically, thus it is syntactically invalid. For more information, see Objects and references.)
Warning
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'?>
Think about global $var;
as a shortcut to $var =& $GLOBALS['var'];
. Thus assigning another reference
to $var
only changes the local variable's reference.
Note:
If you assign a value to a variable with references in a foreach statement, the references are modified too.
Example #3 References and foreach statement
<?php$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 $arr are changed even though it's not a reference! */?>
In other words, the reference behavior of arrays is defined in an element-by-element basis; the reference behavior of individual elements is dissociated from the reference status of the array container.
Pass By Reference
The second thing references do is to pass variables by reference. This is done by making a local variable in a function and a variable in the calling scope referencing the same content. Example:
<?phpfunction foo(&$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.