PHP內存引用

Question

我很長一段時間想知道這個問題，PHP如何處理引用是他們的一個好主意，我不能解釋比使用示例更好，讓我們看看下面的類，然後@ setResult方法的註釋。讓我們想象我們正在使用模型視圖控制器框架，並且我們正在構建基本的AjaxController，到目前爲止我們只有1個操作方法（getUsers）。閱讀評論，我希望我的問題很清楚，PHP如何處理這種情況，並且是我在內存@ setResult docblock中寫了x次的真實情況。

class AjaxController{ 
    private $json = array(
     'result' => array(), 
     'errors' => array(), 
     'debug' => array() 
    ); 

    /** 
    * Adds an error, always displayed to users if any errors. 
    * 
    * @param type $description 
    */ 
    private function addError($description){ 
     $this->json['errors'][] = $description; 
    } 

    /** 
    * Adds an debug message, these are displayed only with DEBUG_MODE. 
    * 
    * @param type $description 
    */ 
    private function addDebug($description){ 
     $this->json['debug'][] = $description; 
    } 

    /** 
    * QUESTION: How does this go in memory? Cause if I use no references, 
    * the array would be 3 times in the memory, if the array is big (5000+) 
    * its pretty much a waste of resources. 
    * 
    * 1st time in memory @ model result. 
    * 2th time in memory @ setResult ($resultSet variable) 
    * 3th time in memory @ $this->json 
    * 
    * @param array $resultSet 
    */ 
    private function setResult($resultSet){ 
     $this->json['result'] = $resultSet; 
    } 

    /** 
    * Gets all the users 
    */ 
    public function _getUsers(){ 
     $users = new Users(); 
     $this->setResult($users->getUsers()); 
    } 

    public function __construct(){ 
     if(!DEBUG_MODE && count($this->json['debug']) > 0){ 
      unset($this->json['debug']); 
     } 

     if(count($this->json['errors']) > 0){ 
      unset($this->json['errors']); 
     } 

     echo json_encode($this->json); 
    } 
} 

另一個簡單的例子：什麼是更好地使用技術答：

function example(){ 
    $latestRequest = $_SESSION['abc']['test']['abc']; 

    if($latestRequest === null){ 
     $_SESSION['abc']['test']['abc'] = 'test'; 
    } 
} 

或技術B：

function example(){ 
    $latestRequest =& $_SESSION['abc']['test']['abc']; 

    if($latestRequest === null){ 
     $latestRequest = 'test'; 
    } 
} 

感謝您的閱讀，並建議:)

Answer 1

在簡而言之：不要使用引用。

寫入PHP副本。考慮：

$foo = "a large string"; 
$bar = $foo; // no copy 
$zed = $foo; // no copy 
$bar .= 'test'; // $foo is duplicated at this point. 
       // $zed and $foo still point to the same string 

當您需要它們提供的功能時，應該只使用引用。即，您需要通過對其的引用來修改原始數組或標量。