ID: 38824
User updated by: tklingenberg at lastflood dot com
Reported By: tklingenberg at lastflood dot com
Status: Wont fix
Bug Type: Feature/Change Request
Operating System: win32
PHP Version: 5.1.6
New Comment:
Accepting a parameter by reference does not mean it's impossible to
check for an uninitalized variable:
<?php
function unintialize_a_var(&$var) {
if (!isset($var)) {
// Why should I uninitalize something uninitialized?
} else {
$var = NULL;
}
}
uninitialize_a_var($uninitialized); // you would expect a NOTICE here,
right?
?>
Anyway your (and mine) example function is useless, and it does not
point to the problem itself afterall.
But if the function is about setting the type of a variable and the
variable is not initialized I would strongly assume to get a NOTICE
about this. Especially in PHP where a variable is created by using the
dollarsign followed by the variabelename.
Would you expect to get a NOTICE here?:
echo $var;
var_dump($var);
intval($var);
I get a notice there. With your arguments you would not expect it here,
right?
Previous Comments:
------------------------------------------------------------------------
[2006-09-14 11:16:54] [EMAIL PROTECTED]
It's just impossible.
settype() function accepts the first parameter by reference.
See fo example:
<?php
function create_a_var(&$var) {
$var = 'created';
}
create_a_var($doesnt_exist); // you would not expect a NOTICE here,
right?
?>
settype() does the same.
------------------------------------------------------------------------
[2006-09-14 11:07:13] tklingenberg at lastflood dot com
Description:
------------
In case a program uses an uninitialized variable passed to settype(),
it should throw a Notice, compareable to echo; intval() and other
variable related functions.
Even if PHP does everything right ($var is a variable you can change
the type of), for the PHP User, it's highly possible she/he made an
error and typed in the wrong variable name. Afterwards the type of the
variable is unchecked, which can lead to even more critical errors.
All this is unnoticed because PHP does not throw a NOTICE.
Reproduce code:
---------------
<?php
$r = settype($var, "float");
?>
Expected result:
----------------
It should throw a Notice
Actual result:
--------------
No Notice is giving that $var is not initialized.
------------------------------------------------------------------------
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Edit this bug report at http://bugs.php.net/?id=38824&edit=1
