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对象通过引用传递。即使赋值运算符似乎没有创建对象的副本。

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

这里是一个简单的，设计的证据：

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，我都会得到'

In both print cases I am getting 'after'

那么，如何通过值传递 $ a 到 set_b（） ，而不是引用？

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

推荐答案

在PHP 5+对象通过引用传递。在PHP 4中，它们通过值传递（这就是为什么它有运行时通过引用，它已被弃用）。

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).

你可以使用PHP5中的'clone'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，而不是你在问题中所说的一切...

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

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