如何在没有参考的情况下复制对象? [英] How to make a copy of an object without reference?

问题描述

有据可查的是，PHP5 OOP 对象是通过引用传递的默认情况下.如果默认情况下是默认设置，那么在我看来，这是一种没有参考的无默认复制方式，怎么办?

It is well documented that PHP5 OOP objects are passed by reference by default. If this is by default, it seems to me there is a no-default way to copy with no reference, how??

function refObj($object){
    foreach($object as &$o){
        $o = 'this will change to ' . $o;
    }

    return $object;
}

$obj = new StdClass;
$obj->x = 'x';
$obj->y = 'y';

$x = $obj;

print_r($x)
// object(stdClass)#1 (3) {
//   ["x"]=> string(1) "x"
//   ["y"]=> string(1) "y"
// }

// $obj = refObj($obj); // no need to do this because
refObj($obj); // $obj is passed by reference

print_r($x)
// object(stdClass)#1 (3) {
//   ["x"]=> string(1) "this will change to x"
//   ["y"]=> string(1) "this will change to y"
// }

在这一点上，我希望$x是原始的$obj，但是当然不是.有什么简单的方法可以做到这一点，还是我必须编写

At this point I would like $x to be the original $obj, but of course it's not. Is there any simple way to do this or do I have to code something like this

推荐答案

<?php
$x = clone($obj);

所以它应该像这样:

<?php
function refObj($object){
    foreach($object as &$o){
        $o = 'this will change to ' . $o;
    }

    return $object;
}

$obj = new StdClass;
$obj->x = 'x';
$obj->y = 'y';

$x = clone($obj);

print_r($x)

refObj($obj); // $obj is passed by reference

print_r($x)

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