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php变量销毁unset的用法

Jun 13, 2016 am 11:18 AM
phpunsetvoidvariableTutorialusageoffreeddestroy

 

php教程变量销毁unset的用法

unset -- 释放给定的变量
描述
void unset ( mixed var [, mixed var [, ...]])


unset() 销毁指定的变量。注意在 PHP 3 中,unset() 将返回 TRUE(实际上是整型值 1),而在 PHP 4 中,unset() 不再是一个真正的函数:它现在是一个语句。这样就没有了返回值,试图获取 unset() 的返回值将导致解析错误。

 

参考php手册:


/* 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教程.


如此便能够说明php 的 unset是如何进行的

先要强调的一点是unset在php中已经不再是一个函数了,既然不是函数,那么就没有了返回值,所以用的时候不能够用unset的返回值来做判断。

其次,在函数中,unset只能销毁局部变量,并不能销毁全局变量,来看下手册的一个例子


function destroy_foo() {
global $foo;
unset($foo);
}

$foo = ‘bar’;
destroy_foo();
echo $foo;
?>

返回的结果为

bar

http://www.bkjia.com/phper/php/37201.htm


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