PHP类不存储引用

Question

如何传递对象构造函数的引用,并允许该对象更新该引用？

class A{
    private $data;
    function __construct(&$d){
        $this->data = $d;
    }
    function addData(){
        $this->data["extra"]="stuff";
    }
}

// Somewhere else
$arr = array("seed"=>"data");
$obj = new A($arr);
$obj->addData();

// I want $arr to contain ["seed"=>"data", "extra"=>"stuff"]
// Instead it only contains ["seed"=>"data"]

Answer 1

您必须将它存储在任何地方作为参考.

function __construct (&$d) {
    $this->data = &$d; // the & here
}