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);

    ?>
    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。

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    用户贡献的备注 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.
    up
    45
    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.
    up
    4
    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!
    ?>
    up
    5
    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;
    ?>
    up
    1
    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.
    up
    2
    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
    up
    -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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