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:
"either create a variable or get a notice."
settype() is not about creating variables. to create a varibale, the
dollarsign ($) is used.
so only the "or get a notice" is left:
I don't want this to be a personal thing, so it infact does not depend
on what I want to get. Additional, I don't want PHP to look into my
head as well.
Maybe its more understandable, when I write down my motivation:
Normally I do use settype() to ensure that an already initialised
variable is set to a specific type. That's all. In that case I wanted
to get a NOTICE. Maybe this really is completety wrong, but using
settype() on an unitialized variables makes no sense, because
uninitalized variables ever contain NULL, so setting their type is
completely useless. So if PHP ecnounters a NULL value inside settype,
it's a strong signal, that something went wrong. So it might be nice of
PHP to drop a NOTICE. It's like using intval() or echo on unitialized
variables.
Previous Comments:
------------------------------------------------------------------------
[2006-09-14 11:45:09] [EMAIL PROTECTED]
It's impossible to detect what you really want to get - either create a
variable or get a notice.
PHP cannot get into your head and read it.
------------------------------------------------------------------------
[2006-09-14 11:36:29] tklingenberg at lastflood dot com
Accepting a parameter by reference does not mean it's impossible to
check for an uninitalized variable:
<?php
function unintialize_a_var(&$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($var);
I get a notice there. With your arguments you would not expect it here,
right?
------------------------------------------------------------------------
[2006-09-14 11:16:54] [EMAIL PROTECTED]
It's just impossible.
settype() function accepts the first parameter by reference.
See fo example:
<?php
function create_a_var(&$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.
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=38824&edit=1
