ID:               49322
 Updated by:       col...@php.net
 Reported By:      anshul at designgrill dot com
 Status:           Bogus
 Bug Type:         Arrays related
 Operating System: All
 PHP Version:      5.2.10
 New Comment:

Again, those are expected results, when you copy an array, its 
elements that are references will be copied as references. It has always
been like that.


Previous Comments:
------------------------------------------------------------------------

[2009-08-21 17:21:28] anshul at designgrill dot com

The example you gave behaves expectedly and actually denotes the bug is
valid. I would request you to carefully go through the two code snippets
I gave.

Even when you copy a value by reference the array, the copy is not
actually made and instead the refcount is increased of the array. PHP
will defer the actual copy until any one of the array is actually
modified. 

Now when you copy the reference to the any array item, and modify using
that reference, a new copy is created and values updated. But if you
keep the reference of the item in a variable before array copying,
modifying this variable will not force PHP to create the actual copy of
the array.

Incorrect behavior at the best.

------------------------------------------------------------------------

[2009-08-21 17:08:22] col...@php.net

Yes, the array itself is copied by value, but not the array content, as
you discovered.

$a = array();
$b = $a; 
$b[] = 2; var_dump(count($a)); // int(0)

------------------------------------------------------------------------

[2009-08-21 17:00:12] anshul at designgrill dot com

Following will give a different result. Just because reference is
copying after assignment.

PHP manual says the assignment operation copies by value for arrays.

<?php
$arr1 = array(1, 2);
$arr2 = $arr1;
$var = &$arr1[0];

$var = 10;
print_r($arr1);
print_r($arr2);
?>

------------------------------------------------------------------------

[2009-08-21 16:54:32] col...@php.net

Sorry, but your problem does not imply a bug in PHP itself.  For a
list of more appropriate places to ask for help using PHP, please
visit http://www.php.net/support.php as this bug system is not the
appropriate forum for asking support questions.  Due to the volume
of reports we can not explain in detail here why your report is not
a bug.  The support channels will be able to provide an explanation
for you.

Thank you for your interest in PHP.

That's expected, arrays have always worked like that.

------------------------------------------------------------------------

[2009-08-21 16:52:29] anshul at designgrill dot com

Description:
------------
If you keep reference to an element of the array before the copy
operation using = operator, you can still modify both the arrays using
the reference copied earlier.

The same doesn't happen if you keep the reference after the copy
operation and try to modify the variable using it.

Reproduce code:
---------------
<?php
$arr1 = array(1, 2);
$var = &$arr1[0];//Keeping the reference
$arr2 = $arr1;//Copy operation
$var = 10;//Will modify both the arrays, bad

print_r($arr1);
print_r($arr2);
?>


Expected result:
----------------
Array
(
    [0] => 10
    [1] => 2
)
Array
(
    [0] => 1
    [1] => 2
)


Actual result:
--------------
Array
(
    [0] => 10
    [1] => 2
)
Array
(
    [0] => 10
    [1] => 2
)



------------------------------------------------------------------------


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