Simple look how PHP Reference works
<?php
/* Imagine this is memory map
______________________________
|pointer | value | variable |
-----------------------------------
| 1 | NULL | --- |
| 2 | NULL | --- |
| 3 | NULL | --- |
| 4 | NULL | --- |
| 5 | NULL | --- |
------------------------------------
Create some variables */
$a=10;
$b=20;
$c=array ('one'=>array (1, 2, 3));
/* Look at memory
_______________________________
|pointer | value | variable's |
-----------------------------------
| 1 | 10 | $a |
| 2 | 20 | $b |
| 3 | 1 | $c['one'][0] |
| 4 | 2 | $c['one'][1] |
| 5 | 3 | $c['one'][2] |
------------------------------------
do */
$a=&$c['one'][2];
/* Look at memory
_______________________________
|pointer | value | variable's |
-----------------------------------
| 1 | NULL | --- | //value of $a is destroyed and pointer is free
| 2 | 20 | $b |
| 3 | 1 | $c['one'][0] |
| 4 | 2 | $c['one'][1] |
| 5 | 3 | $c['one'][2] ,$a | // $a is now here
------------------------------------
do */
$b=&$a; // or $b=&$c['one'][2]; result is same as both "$c['one'][2]" and "$a" is at same pointer.
/* Look at memory
_________________________________
|pointer | value | variable's |
--------------------------------------
| 1 | NULL | --- |
| 2 | NULL | --- | //value of $b is destroyed and pointer is free
| 3 | 1 | $c['one'][0] |
| 4 | 2 | $c['one'][1] |
| 5 | 3 |$c['one'][2] ,$a , $b | // $b is now here
---------------------------------------
next do */
unset($c['one'][2]);
/* Look at memory
_________________________________
|pointer | value | variable's |
--------------------------------------
| 1 | NULL | --- |
| 2 | NULL | --- |
| 3 | 1 | $c['one'][0] |
| 4 | 2 | $c['one'][1] |
| 5 | 3 | $a , $b | // $c['one'][2] is destroyed not in memory, not in array
---------------------------------------
next do */
$c['one'][2]=500; //now it is in array
/* Look at memory
_________________________________
|pointer | value | variable's |
--------------------------------------
| 1 | 500 | $c['one'][2] | //created it lands on any(next) free pointer in memory
| 2 | NULL | --- |
| 3 | 1 | $c['one'][0] |
| 4 | 2 | $c['one'][1] |
| 5 | 3 | $a , $b | //this pointer is in use
---------------------------------------
lets tray to return $c['one'][2] at old pointer an remove reference $a,$b. */
$c['one'][2]=&$a;
unset($a);
unset($b);
/* look at memory
_________________________________
|pointer | value | variable's |
--------------------------------------
| 1 | NULL | --- |
| 2 | NULL | --- |
| 3 | 1 | $c['one'][0] |
| 4 | 2 | $c['one'][1] |
| 5 | 3 | $c['one'][2] | //$c['one'][2] is returned, $a,$b is destroyed
--------------------------------------- ?>
I hope this helps.
リファレンスの解除
リファレンスを解除することは、ちょうど変数名と変数の内容の結合を 解除することに相当します。これは、変数の内容が破棄されることを 意味しません。例えば、
<?php
$a = 1;
$b =& $a;
unset($a);
?>ここでも、Unix の unlink コールと類似したものと 考えると便利です。
add a note
User Contributed Notes 7 notes
ojars26 at NOSPAM dot inbox dot lv ¶
16 years ago
sony-santos at bol dot com dot br ¶
17 years ago
<?php
//if you do:
$a = "hihaha";
$b = &$a;
$c = "eita";
$b = $c;
echo $a; // shows "eita"
$a = "hihaha";
$b = &$a;
$c = "eita";
$b = &$c;
echo $a; // shows "hihaha"
$a = "hihaha";
$b = &$a;
$b = null;
echo $a; // shows nothing (both are set to null)
$a = "hihaha";
$b = &$a;
unset($b);
echo $a; // shows "hihaha"
$a = "hihaha";
$b = &$a;
$c = "eita";
$a = $c;
echo $b; // shows "eita"
$a = "hihaha";
$b = &$a;
$c = "eita";
$a = &$c;
echo $b; // shows "hihaha"
$a = "hihaha";
$b = &$a;
$a = null;
echo $b; // shows nothing (both are set to null)
$a = "hihaha";
$b = &$a;
unset($a);
echo $b; // shows "hihaha"
?>
I tested each case individually on PHP 4.3.10.
lazer_erazer ¶
18 years ago
Your idea about unsetting all referenced variables at once is right,
just a tiny note that you changed NULL with unset()...
again, unset affects only one name and NULL affects the data,
which is kept by all the three names...
<?php
$a = 1;
$b =& $a;
$b = NULL;
?>
This does also work!
<?php
$a = 1;
$b =& $a;
$c =& $b;
$b = NULL;
?>
donny at semeleer dot nl ¶
18 years ago
Here's an example of unsetting a reference without losing an ealier set reference
<?php
$foo = 'Bob'; // Assign the value 'Bob' to $foo
$bar = &$foo; // Reference $foo via $bar.
$bar = "My name is $bar"; // Alter $bar...
echo $bar;
echo $foo; // $foo is altered too.
$foo = "I am Frank"; // Alter $foo and $bar because of the reference
echo $bar; // output: I am Frank
echo $foo; // output: I am Frank
$foobar = &$bar; // create a new reference between $foobar and $bar
$foobar = "hello $foobar"; // alter $foobar and with that $bar and $foo
echo $foobar; //output : hello I am Frank
unset($bar); // unset $bar and destroy the reference
$bar = "dude!"; // assign $bar
/* even though the reference between $bar and $foo is destroyed, and also the
reference between $bar and $foobar is destroyed, there is still a reference
between $foo and $foobar. */
echo $foo; // output : hello I am Frank
echo $bar; // output : due!
?>
smcbride at msn dot com ¶
2 years ago
A little quirk on unset() when using references that may help someone.
If you want to delete the element of a reference to an array, you need to have the reference point to the parent of the key that you want to delete.
<?php
$arr = array('foo' => array('foo_sub1' => 'hey', 'foo_sub2' => 'you'), 'bar' => array('bar_sub1' => 'good', 'bar_sub2' => 'bye'));
$parref = &$arr['foo'];
$childref = &$parref['foo_sub1'];
unset($childref); // this will simply unset the reference to child
unset($parref['foo_sub1']); // this will actually unset the data in $arr;
$parref['foo_sub1'] = NULL; // this will set the element to NULL, but not delete it. If you run it after unset(), it add the key back and set it to NULL
?>
This is nice to use for passing something dynamically to a function by reference without copying the entire array to the function, but you want to do some maintenance on the array.
frowa at foxmail dot com ¶
3 years ago
it's my way to remember.
<?php
// the var $a is point to the value 1, as a line connect to value 1
$a = 1;
// the var $b point to the value which the var $a point to, as a new line connect to value 1
$b =& $a;
// cut the line of the var $a to value 1,now $a is freedom,it's nothing point to. so the value of $a is null
unset($a);
?>
$a--------> 1
↑
|
|
$b
libi ¶
18 years ago
clerca at inp-net dot eu dot org
"
If you have a lot of references linked to the same contents, maybe it could be useful to do this :
<?php
$a = 1;
$b = & $a;
$c = & $b; // $a, $b, $c reference the same content '1'
$b = NULL; // All variables $a, $b or $c are unset
?>
"
------------------------
NULL will not result in unseting the variables.
Its only change the value to "null" for all the variables.
becouse they all points to the same "part" in the memory.
