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call_user_func_array这个函数,该如何处理

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Release: 2016-06-13 13:39:11
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call_user_func_array这个函数
是不是只能使用在没有构造函数的类,比如:

PHP code
<!--

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  class service {
      
    private $self_id ;
    
    /* 构造 */
    public function __construct($a) {
           /* ……
            */
    }

    public function A($b) {
    }

    }

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现在要想用call_user_func_array() 这个函数调用service类的function A() 该怎么写?构造函数的参数怎么办?

------解决方案--------------------
通常实现你这个功能会在初此类中 _call 来实现,传过一的参数不确定,但调用的参数会转化为 数组的。
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