PHP 对象分配与克隆 [英] PHP Object Assignment vs Cloning

问题描述

我知道这在 php 文档中有介绍，但我对这个问题感到困惑.

I know this is covered in the php docs but I got confused with this issue .

来自 php 文档:

$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);
?>

上面的例子会输出:

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

好的，所以我看到 $assigned survived 原始对象 ($instance) 被分配给 null，所以很明显 $assigned 不是引用而是 $instance 的副本.

OK so I see that $assigned survived the original object ($instance) being assigned to null, so obviously $assigned isn't a reference but a copy of $instance.

那么

 $assigned = $instance 

和

 $assigned = clone $instance

推荐答案

对象是内存中的抽象数据.变量总是在内存中保存对这个数据的引用.想象一下，$foo = new Bar 在内存中的某处创建了一个 Bar 的对象实例，为它分配了一些 id #42 和 $foo 现在持有这个 #42 作为对这个对象的引用.将此引用分配给其他变量通过引用或通常与任何其他值相同.许多变量可以保存此引用的副本，但都指向同一个对象.

Objects are abstract data in memory. A variable always holds a reference to this data in memory. Imagine that $foo = new Bar creates an object instance of Bar somewhere in memory, assigns it some id #42, and $foo now holds this #42 as reference to this object. Assigning this reference to other variables by reference or normally works the same as with any other values. Many variables can hold a copy of this reference, but all point to the same object.

clone 显式地创建对象本身的副本，而不仅仅是指向对象的引用的副本.

clone explicitly creates a copy of the object itself, not just of the reference that points to the object.

$foo = new Bar;   // $foo holds a reference to an instance of Bar
$bar = $foo;      // $bar holds a copy of the reference to the instance of Bar
$baz =& $foo;     // $baz references the same reference to the instance of Bar as $foo

只是不要将 =& 中的引用"与对象标识符中的引用"混淆.

Just don't confuse "reference" as in =& with "reference" as in object identifier.

$blarg = clone $foo;  // the instance of Bar that $foo referenced was copied
                      // into a new instance of Bar and $blarg now holds a reference
                      // to that new instance

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