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Home/ Questions/Q 3342318
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T00:49:28+00:00 2026-05-18T00:49:28+00:00

It is well documented that PHP5 OOP objects are passed by reference by default.

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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

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T00:49:28+00:00Added an answer on May 18, 2026 at 12:49 am
    <?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)
    
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