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php – How to make a copy of an object without reference?

Posted by: admin April 23, 2020 Leave a comment

Questions:

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"
// }

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

How to&Answers:
<?php
$x = clone($obj);

So it should read 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)

Answer:

To make a copy of an object, you need to use object cloning.

To do this in your example, do this:

$x = clone $obj;

Note that objects can define their own clone behaviour using __clone(), which might give you unexpected behaviour, so bear this in mind.