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.
PHP - Manual: 取消引用
2024-11-15
取消引用
当 unset 一个引用，只是断开了变量名和变量内容之间的绑定。这并不意味着变量内容被销毁了。例如：
<?php
$a = 1;
$b =& $a;
unset($a);
?>
再拿这个和 Unix 的 unlink 调用来类比一下可能有助于理解。
add a note
User Contributed Notes 7 notes
ojars26 at NOSPAM dot inbox dot lv ¶
14 years ago
sony-santos at bol dot com dot br ¶
15 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 ¶
15 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 ¶
15 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 ¶
6 months 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 ¶
1 year 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 ¶
16 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.
官方地址：https://www.php.net/manual/en/language.references.unset.php
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