如何在 PHP 中创建对象的副本? [英] How do I create a copy of an object in PHP?

问题描述

似乎在 PHP 中对象是通过引用传递的.甚至赋值运算符似乎也没有创建 Object 的副本.

It appears that in PHP objects are passed by reference. Even assignment operators do not appear to be creating a copy of the Object.

这是一个简单的、人为的证明:

Here's a simple, contrived proof:

<?php

class A {
    public $b;
}


function set_b($obj) { $obj->b = "after"; }

$a = new A();
$a->b = "before";
$c = $a; //i would especially expect this to create a copy.

set_b($a);

print $a->b; //i would expect this to show 'before'
print $c->b; //i would ESPECIALLY expect this to show 'before'

?>

在这两种打印情况下，我都得到了之后"

In both print cases I am getting 'after'

那么，我如何将 $a 通过值而不是引用传递给 set_b()?

So, how do I pass $a to set_b() by value, not by reference?

推荐答案

在 PHP 5+ 中，对象是通过引用传递的.在 PHP 4 中，它们是按值传递的(这就是为什么它在运行时通过引用传递，这已被弃用).

In PHP 5+ objects are passed by reference. In PHP 4 they are passed by value (that's why it had runtime pass by reference, which became deprecated).

您可以使用 PHP5 中的克隆"运算符来复制对象:

You can use the 'clone' operator in PHP5 to copy objects:

$objectB = clone $objectA;

此外，它只是通过引用传递的对象，而不是您在问题中所说的所有内容......

Also, it's just objects that are passed by reference, not everything as you've said in your question...

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