title: Deprecated: Assigning the return value of new by reference is deprecated in D:\xampp\php\PEAR\Config.php on line 80
body: <pre><code>    &lt;?php
  /**
  * This class is sort of factory class that is responsible for loading
  * classes, it check if this class is not defined then it includes the file
  * So developer don't need to worry about including that file
  * @author Haafiz
  */
  class load{
  
      public static $app_path= APP_PATH;
      public static $model_path=MODEL_PATH;
  
  
      /*
      * @param string $model_name &lt;&gt;Name of class(model) that is required to instantiate/load&lt;/p&gt;
      * @param bool $continue_on_error  this decide whether to have fatal error or continue on error loading
      * $return object 
      */
      public static function model($model_name,$conitnue_on_error=0){
          if(!class_exists($model_name)){
              $model_filename= strtolower($model_name).".php";
              try{
                  include self::$model_path.$model_filename;
              }
              catch(Exception $e){
                  if(!$continue_on_error){
                  die($e);
                  }
              }
              $model=new $model_name();
              return $model;
          }       
      }
  }
  
  ?&gt;
  </code></pre>
  
  <p>In above code have to face following error. Some people says in some other threads that problem is in using <code>&amp;</code> , I am not using that. So what actually is prolem in my case? All seems to doing every thing correctly. Saw some other threads but didn't find any solution. So please if other people understand any thing by it.
  thanks</p>