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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);
?>

不会 unset $b，只是 $a。

再拿这个和 Unix 的 unlink 调用来类比一下可能有助于理解。

User Contributed Notes

ojars26 at NOSPAM dot inbox dot lv 03-May-2008 07:41


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.

sony-santos at bol dot com dot br 05-Feb-2007 03:56


<?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 05-Sep-2006 04:02


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 13-Jul-2006 07:10


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!

?>

libi 24-Jan-2006 12:20


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.

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