php – 对象分配与引用

有一个声明：

When assigning an already created instance of a class to a new variable,the new variable will access the same instance as the object that was assigned. This behavior is the same when passing instances to a function. A copy of an already created object can be made by cloning it.

我假设这是默认情况下对象通过引用传递的状态,因此如果要创建真实副本,则应该是clone. (PHP中没有浅拷贝.是的,默认情况下有一个克隆.)

请考虑以下示例(从上面的链接复制)：

<?php

$instance = new SimpleClass();

$assigned   =  $instance;
$reference  =& $instance;

$instance->var = '$assigned will have this value';

$instance = null; // $instance and $reference become null

var_dump($instance);
var_dump($reference);
var_dump($assigned);

?>

正如在那里被告知的那样,输出如下：

NULL
NULL
object(SimpleClass)#1 (1) {
   ["var"] => string(30) "$assigned will have this value"
}

我不明白.

如果$assigned = $instance;默认情况下是对象的引用(别名)赋值,而不是$assigned仍然是SimpleClass的一个对象,它保存带有该字符串的$var属性,而NULL被分配给$instance.