ID: 46885
Updated by: [email protected]
Reported By: Chowarmaan at gmail dot com
-Status: Open
+Status: Feedback
Bug Type: Scripting Engine problem
Operating System: Windows 2000/XP
PHP Version: 5.2.8
New Comment:
How about you comment out the latter call and let the script work
like it does?
Previous Comments:
------------------------------------------------------------------------
[2008-12-16 21:26:23] Chowarmaan at gmail dot com
Description:
------------
Calling a function in a class that accepts a variable by reference,
then checks the variable and creates an object for it will not allow the
object to be returned outside of the function.
Reproduce code:
---------------
<?php
class TEST_CLASS
{
function __construct() { }
function GetResponse(&$Response, $Timeout = 0)
{
if (!($Response instanceof TEST_CLASS2 ))
$Response = new TEST_CLASS2();
$Response->SetText_('Testing Complete');
return TRUE;
}
}
class TEST_CLASS2
{
public $Variable;
public function SetText_($s)
{
$this->Variable = $s;
}
}
$Test = new TEST_CLASS();
$Test->GetResponse($Response, 100); // Fails
$Test->GetResponse(&$Response, 100); // Works
echo $Response->Variable . "\n";
?>
Expected result:
----------------
$Response should be of type TEST_CLASS2 and the echo should return the
text:
Testing Complete
Actual result:
--------------
$Response is a NULL
------------------------------------------------------------------------
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Edit this bug report at http://bugs.php.net/?id=46885&edit=1