Question

我有这堂课：

class Foo {

   private $_data;

   public function __construct(array $data){
       $this->_data = $data;
   }

   public function __get($name){
       $getter = 'get'.$name;
       if(method_exists($this, $getter)){
           return $this->$getter();
       }

       if(array_key_exists($name,$this->_data)){
           return $this->_data[$name];
       }

       throw new Exception('Property '.get_class($this).'.'.$name.' is not available');
   }

   public function getCalculated(){
      return null;
   }
}

getCalculated()表示计算的属性。

现在，如果我尝试以下内容：

 $foo = new Foo(['related' => []])
 $foo->related[] = 'Bar'; // Indirect modification of overloaded property has no effect

 $foo->calculated; // ok

但如果我将__get()签名更改为&__get($name)，我会：

 $foo = new Foo(['related' => []])
 $foo->related[] = 'Bar'; // ok

 $foo->calculated; // Only variables should be passed by reference

我非常希望通过引用返回$data元素，并在__get()中按值获取getter。这可能吗？

Answer 1

如错误消息所示，您需要从getter中返回一个变量：

class Foo {

   private $_data;

   public function __construct(array $data){
       $this->_data = $data;
   }

   public function  &__get($name){
       $getter = 'get'.$name;
       if(method_exists($this, $getter)){
           $val = $this->$getter();  // <== here we create a variable to return by ref
           return $val;
       }

       if(array_key_exists($name,$this->_data)){
           return $this->_data[$name];
       }

       throw new Exception('Property '.get_class($this).'.'.$name.' is not available');
   }

   public function getCalculated(){
      return null;
   }
}

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1 个答案: