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。

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
//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.

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!
?>

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; 
?>

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.