Thanks for help Daniel Clark. I know i must use isset function but that was
not the issue. But thanks anyway:)
Curt Zirzow thanks for your reply also. But that has nothing to do with
reference. If you use reference or not.. it still would not result in an
error (hee... I want an error :P)
Matt Matijevich thanks for your reply as well :)
My function is already declared something like this:
function testX($testArr)
{
$arr_test = $testArr;
return $arr_test;
}
Even if you send a reference parameter (&$testArr).. the outcome will not
result in an error.
I guess somehow PHP does something to the function argument if it does not
exist: creating it and give it the NULL value. But this should at least give
us a warning because it could result in unwanted code!
But i dunno know this for sure. Could anybody help me on this?
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
