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php的引用返回和取消引用

Jun 13, 2016 am 10:54 AM
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一、引用返回

引用返回用在当想用函数找到引用应该被绑定在哪一个变量上面时。不要用返回引用来增加性能,引擎足够聪明来自己进行优化。仅在有合理的技术原因时才返回引用!要返回引用,使用此语法:

[php]  

class foo {  

    public $value = 42;  

  

    public function &getValue() {  

        return $this->value;  

    }  

}  

  

$obj = new foo;  

$myValue = &$obj->getValue(); // $myValue is a reference to $obj->value, which is 42.  

$obj->value = 2;  

echo $myValue;                // prints the new value of $obj->value, i.e. 2.  

?>  

 

以上是PHP Manual给出的解释并表示太好理解

[php]  

    function &test(){  

        static $b = 0;  //声明一个静态变量  

        $b = $b+1;  

        echo $b."
";  

        return $b;  

    }  

   $a = test();  //输出 $b 的值为:1  

   $a = 5;  

   $a = test();  //输出 $b 的值为:2  

    

   $a = &test();  //输出 $b 的值为:3  **注意**  

   $a = 5;    //$b的值变为了5  

   $a = test();  //输出 $b 的值为:6  **注意**  

   ?>  

 

 

$a = test() 虽然说函数定义的时候,是引用返回方式,但是如果采用这种普通形势调用函数,那它的作用也就和普通的函数一样,所以看结果就是1、2

 

$a = &test() 这种调用方式就是引用返回,就类似于 $a = &$b ,然后第二句又把$a = 5,那就是等于将变量$b = 5,最后一句得到的6也就很容易理解了!

 

和参数传递不同,这里必须在两个地方都用 & 符号——指出返回的是一个引用,而不是通常的一个拷贝,同样也指出 $a 是作为引用的绑定,而不是通常的赋值。www.2cto.com

 

Note: 如果试图这样从函数返回引用:return ($this->value);,这将不会起作用,因为在试图返回一个表达式的结果而不是一个引用的变量。只能从函数返回引用变量——没别的方法。如果代码试图返回一个动态表达式或 new 运算符的结果,自 PHP 4.4.0 和 PHP 5.1.0 起会发出一条 E_NOTICE 错误。

 

二、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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