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Home Backend Development PHP Tutorial PHP提升性能 unset销毁变量并释放内存

PHP提升性能 unset销毁变量并释放内存

Jun 20, 2016 pm 01:01 PM
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PHP unset销毁变量并释放内存

PHP的unset()函数用来清除、销毁变量,不用的变量,我们可以用unset()将它销毁。但是某些时候,用unset()却无法达到销毁变量占用的内存!我们先看一个例子:

<?php $s=str_repeat('1',255); //产生由255个1组成的字符串
$m=memory_get_usage(); //获取当前占用内存
unset($s);
$mm=memory_get_usage(); //unset()后再查看当前占用内存
echo $m-$mm;
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最后输出unset()之前占用内存减去unset()之后占用内存,如果是正数,那么说明unset($s)已经将$s从内存中销毁(或者说,unset()之后内存占用减少了),可是我在PHP5和windows平台下,得到的结果是:-48。这是否可以说明,unset($s)并没有起到销毁变量$s所占用内存的作用呢?我们再作下面的例子:

<?php $s=str_repeat('1',256); //产生由256个1组成的字符串
$m=memory_get_usage(); //获取当前占用内存
unset($s);
$mm=memory_get_usage(); //unset()后再查看当前占用内存
echo $m-$mm;
Copy after login

这个例子,和上面的例子几乎相同,唯一的不同是,$s由256个1组成,即比第一个例子多了一个1,得到结果是:224。这是否可以说明,unset($s)已经将$s所占用的内存销毁了?

通过上面两个例子,我们可以得出以下结论:结论一、unset()函数只能在变量值占用内存空间超过256字节时才会释放内存空间。

那么是不是只要变量值超过256,使用unset就可以释放内存空间呢?我们再通过一个例子来测试一下:

<?php $s=str_repeat('1',256); //这和第二个例子完全相同
$p=&$s;
$m=memory_get_usage();
unset($s); //销毁$s
$mm=memory_get_usage();
echo $p.'<br />';
echo $m-$mm;
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刷新页面,我们看到第一行有256个1,第二行是-48,按理说我们已经销毁了$s,而$p只是引用$s的变量,应该是没有内容了,另外,unset($s)后内存占用却比unset()前增加了!现在我们再做以下的例子:

<?php $s=str_repeat('1',256); //这和第二个例子完全相同
$p=&$s;
$m=memory_get_usage();
$s=null; //设置$s为null
$mm=memory_get_usage();
echo $p.'<br />';
echo $m-$mm;
Copy after login

现在刷新页面,我们看到,输出$p已经是没有内容了,unset()前后内存占用量之差是224,即已经清除了变量占用的内存。本例中的$s=null也可以换成unset(),如下:

<?php $s=str_repeat('1',256); //这和第二个例子完全相同
$p=&$s;
$m=memory_get_usage();
unset($s); //销毁$s
unset($p);
$mm=memory_get_usage();
echo $p.'<br />';
echo $m-$mm;
Copy after login

我们将$s和$p都使用unset()销毁,这时再看内存占用量之差也是224,说明这样也可以释放内存。那么,我们可以得到另外一条结论:结论二、只有当指向该变量的所有变量(如引用变量)都被销毁后,才会释放内存。

相信经过本文的例子后,大家应该对unset()有所了解了,最起码,本人用unset()也是为了在变量不起作用时,释放内存。


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