Edit report at https://bugs.php.net/bug.php?id=30288&edit=1
ID: 30288
Updated by: ni...@php.net
Reported by: Jones at miliz dot org
Summary: Problems making a variable a reference which was
passed by reference
-Status: Open
+Status: Wont fix
Type: Feature/Change Request
-Package: Feature/Change Request
+Package: *General Issues
Operating System: Linux
PHP Version: 4.3.9
Block user comment: N
Private report: N
New Comment:
Assigning by-reference to a variable that already is a reference will break the
existing reference. What you want is write "$second = $source" instead of
"$second =& $source". This will fix your issue.
Previous Comments:
------------------------------------------------------------------------
[2004-09-30 11:04:51] Jones at miliz dot org
Description:
------------
Making a variable, which was passed by reference, a reference to another
variable inside the function won't work outside this function
Reproduce code:
---------------
<?php
function &testReference(&$first, &$second) {
$source = 'my Test';
$first = $source;
$second =& $source;
var_dump($second);
return $second;
}
$ref3 =& testReference($ref1, $ref2);
var_dump($ref1);
var_dump($ref2);
var_dump($ref3);
?>
Expected result:
----------------
string(7) "my Test"
string(7) "my Test"
string(7) "my Test"
string(7) "my Test"
Actual result:
--------------
string(7) "my Test"
string(7) "my Test"
NULL
string(7) "my Test"
------------------------------------------------------------------------
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Edit this bug report at https://bugs.php.net/bug.php?id=30288&edit=1
