PHP: Destruindo referências

Destruindo referências

Quando uma referência é quebrada, ela apenas para de fazer o apontamento entre o nome da variável e o conteúdo. Mas isto não significa que o conteúdo da variável é destruído. Por exemplo:

<?php

$a = 1;
$b =& $a;
unset ($a);

?>

não apaga $b, apenas $a.

Novamente, é mais fácil pensar em analogia ao comando UNIX unlink.

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Notas Enviadas por Usuários (em inglês) 7 notes

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442

ojars26 at NOSPAM dot inbox dot lv ¶

16 years ago

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.

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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.

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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!
?>

down

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; 
?>

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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.

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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

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-3

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.

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