[PHP-DEV] PHP 4.0 Bug #8139 Updated: NULL and isset From: stas <email protected>
Date: 12/14/00

ID: 8139
Updated by: stas
Reported By: richard.heyes <email protected>
Old-Status: Open
Status: Closed
Bug Type: Scripting Engine problem
Assigned To:
Comments:

>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.

Previous Comments:
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[2000-12-07 04:22:20] richard.heyes <email protected>
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);
?>
)

---------------------------------------------------------------------------

[2000-12-06 13:18:04] joey <email protected>
The question is: should passing an unset variable by reference
generate a warning?

---------------------------------------------------------------------------

[2000-12-06 12:57:39] richard.heyes <email protected>
<?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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Full Bug description available at: http://bugs.php.net/?id=8139 

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