ID: 49322
User updated by: anshul at designgrill dot com
Reported By: anshul at designgrill dot com
Status: Bogus
Bug Type: Arrays related
Operating System: All
PHP Version: 5.2.10
New Comment:
Moreover, variables, which are references, when copied are not copied
as reference and instead a true copy is made. Array elements should not
be treated differently.
Previous Comments:
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[2009-08-21 21:27:12] anshul at designgrill dot com
Ok.
But it is not obvious from the documentation. A cautionary statement
should be included in there.
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[2009-08-21 21:09:20] col...@php.net
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.
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[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.
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[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)
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[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);
?>
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