ID: 41242
Updated by: [EMAIL PROTECTED]
Reported By: laurynas at by dot lt
Status: Bogus
Bug Type: Arrays related
Operating System: Linux 2.6.17.7
PHP Version: 5.2.1
New Comment:
No, the behavior is undefined (and currently executed after the
function call - but that doesn't make it defined - might change anytime)
Previous Comments:
------------------------------------------------------------------------
[2007-04-30 16:13:03] laurynas at by dot lt
Agree, this is expresion, but the expression is executed before
function is called, right? So the variable should be defined *before*
function takes variable as a parameter. Then why it works in PHP 4.x?
------------------------------------------------------------------------
[2007-04-30 16:06:22] [EMAIL PROTECTED]
"$user = array()" is not a variable, but an expression.
------------------------------------------------------------------------
[2007-04-30 16:04:03] laurynas at by dot lt
Description:
------------
Array values are destroyed after using it as reference in function call
defining in call dirrectly as array. If given array is constructed
before function call in the same parameters line, PHP returns nothing,
but really shoud return a value associated to array.
Reproduce code:
---------------
<?php
// this is our defined function
// as you can see the second parameter is reference
function someFunc($param1, & $array) {
// so expecting $array as array we
// put some value in it
$array[0] = 'some value';
}
// now call the function, please note, that array $user
// is created as array type *outside* function
someFunc(null, $user = array());
// now let's check returned value
print_r($user);
?>
Expected result:
----------------
Array
(
[0] => some value
)
Actual result:
--------------
Array
(
)
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=41242&edit=1
