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Bug #8139 NULL and isset
Submitted: 2000-12-06 12:57 UTC Modified: 2000-12-14 05:07 UTC
From: richard dot heyes at heyes-computing dot net Assigned:
Status: Closed Package: Scripting Engine problem
PHP Version: 4.0.3pl1 OS: Linux
Private report: No CVE-ID: None
View Developer Edit
 [2000-12-06 12:57 UTC] richard dot heyes at heyes-computing dot net 
<?php
function foo(&$bar){
    return TRUE;
}

foo($bar);

print($bar);

if(!isset($bar))
    print('Bar is not set!');
?>

With notices turned on (error_reporting) the call to print($bar) does not throw an error. However the bit below it runs the print() call. So if it's not set, the first print should throw an error regarding the unset $bar. Seems very contradictory. FWIW the type of $bar after the call to foo() is NULL.

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 [2000-12-06 13:18 UTC] joey@php.net 
The question is: should passing an unset variable by reference
generate a warning?
 [2000-12-07 04:22 UTC] richard dot heyes at heyes-computing dot net 
No idea. Though I do know of people who use that method to set variables, so their answer would be a resounding no. :)

(EG:
<?php
function mconnect(&$link){
    $link = mysql_connect();
}

mconnect($db);
?>
)
 [2000-12-14 05:07 UTC] stas@php.net 
From current Zend code I see that no operation that should
write variable (including passing by reference) won't
generate "undefined" waring. This seems logical - if you
write it, why you care if it wasn't set before?

So I'd close it, since it's not a bug, but intended
behaviour. If you disagree, submit a feature request.
 
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