All objects added to an array via looped array_push() calls have property value unintentionally set by final declaration

Question

This is the example code:

//$pieces is an stdClass object which has 4 elements the foreach loops through
$arr = array();
foreach($pieces as $piece)
{
     $piece->value = 1;
     array_push($arr, $piece);

     $piece->value = 3;
     array_push($arr, $piece);
}

The problem with that is it doesn't use the first array_push, just like it wasnt there, in the results I got:

Array
(
    [0] => stdClass Object
        (
             [piece] = 3
        )
    [1] => stdClass Object
        (
             [piece] = 3
        )
    [2] => stdClass Object
        (
             [piece] = 3
        )
    [3] => stdClass Object
        (
             [piece] = 3
        )
)

While there should be additional 4 keys with the [piece] = 1. Am I doing something wrong?

Tip: Have you considered using $array[] = $value vs array_push for single items? — Marty
– Marty, Commented Feb 14, 2014 at 0:52
Of course, it's just nicer to read, faster to type and even more efficient (on a tiny scale). — Marty
– Marty, Commented Feb 14, 2014 at 1:07

marionebl · Accepted Answer · 2014-02-14 00:53:59Z

1

You'll have to clone the $piece object, your code currently saves references to $piece into $arr. This snippet assumes you need actual copies for both $piece variants in your array.

$arr = array();
foreach($pieces as $piece)
{
     $first_clone = clone $piece;
     $first_clone->value = 1;
     array_push($arr, $first_clone);

     $second_clone = clone $piece;
     $second_clone->value = 3;
     array_push($arr, $second_clone);
}

answered Feb 14, 2014 at 0:53

marionebl

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2 Comments

Lucas Over a year ago

I guess the second clone is not needed?

marionebl Over a year ago

Depends on your use case. If you need all the objects in $arr to be independent from $pieces you need both clones.

Niet the Dark Absol · Accepted Answer · 2014-02-14 00:48:07Z

1

Objects are always references. You would need to clone the object before trying to use it as if it were two different things.

answered Feb 14, 2014 at 0:48

Niet the Dark Absol

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