How to copy an object without a reference?

Question

It is well documented that PHP5 OOP objects are passed by reference by default. If this is the default, it seems to me that there is a non-default way to copy without a reference, how about that? ?

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"
// }

At this point I expected $x to be the original $obj, but of course it isn't. Is there any easy way to do this or do I have to write code like this

Answer 1

To copy an object you need to use Object Clone一个>.

To do this in your example:

$x = clone $obj;

Note that objects can define their own clone behavior using __clone(), which may give you unexpected behavior, so please keep this in mind.

Answer 2

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

So it should be like this:

<?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)