关于call_user_func_array 和 引用 的一段代码 不懂求解释
class PbObject{<br /> <br /> var $params;<br /> var $fontFace = 'incite.ttf';<br /> <br /> function Object() {<br /> $args = func_get_args();<br /> if (method_exists($this, '__destruct')) {<br /> register_shutdown_function (array(&$this, '__destruct'));<br /> }<br /> call_user_func_array(array(&$this, '__construct'), $args);<br /> }<br /> <br /> function __construct(){} <br /> <br /> function toString() {<br /> $class = get_class($this);<br /> return $class;<br /> }<br /> }
请问这段代码中的 call_user_func_array(array(&$this, '__construct'), $args); 为什么$this前面加引用符&? 这样的代码我一直没有搞懂过。 不加&不也是一样能调用本身的 __construct 方法? 加了有什么作用?
------解决方案--------------------
当 call_user_func_array 调用的函数是类的方法时,需要用数组同时传递 类(对象)和 方法名
你的 $this 就是对象了
在 php4 中,对象默认以值传递,如果没有引用声明(&),那么实际上是在另一个对象上执行的,如果方法中修改了类属性,那么这一修改将不能作用到调用者本身
在 php5 中,对象都是以引用传递,所以就没有必要冠以 & 了
你的代码中用 var 声明类属性,这显然是 php4 风格的
php4 的类没有析构函数,所以代码用 register_shutdown_function 注册了一个

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