ID: 32789
Updated by: [EMAIL PROTECTED]
Reported By: php at thoftware dot de
-Status: Open
+Status: Bogus
Bug Type: Arrays related
Operating System: *
PHP Version: 4.3.11
New Comment:
This works:
<?php
class foobar {
var $cache = array();
function &cache($j) {
if (!isset($this->cache[$j])) {
$this->cache[$j] = array(
'wert' => $j,
'text' => 'value: '.$j.'<br>',
'time' => 'set at '.date('h:i:s').'<br>',
);
}
return($this->cache[$j]);
}
}
$foobar = new foobar();
$v=$foobar->cache(3);
array_pop($foobar->cache(3));
var_dump($foobar->cache(3));
?>
Ask further support questions on [email protected]
Previous Comments:
------------------------------------------------------------------------
[2005-04-21 17:34:24] php at thoftware dot de
And another thing: As the error occures every time when using a plain
function instead of a method (no matter if the functions result is
assigned to a variable prior or not), I thought it should be an error
giving the result of a function to array_pop(). Or is there any
difference between calling a method or calling a function (related to
this topic, I mean).
------------------------------------------------------------------------
[2005-04-21 17:28:25] php at thoftware dot de
So then pls explain, why this happens:
Reproduce code:
---------------
class foobar {
var $cache = array();
function cache($j) {
if (!isset($this->cache[$j])) {
$this->cache[$j] = array(
'wert' => $j,
'text' => 'value: '.$j.'<br>',
'time' => 'set at '.date('h:i:s').'<br>',
);
}
return($this->cache[$j]);
}
}
$foobar =& new foobar();
$v = $foobar->cache(3);
echo $v['time'];
echo array_pop($foobar->cache(3));
Expected result:
----------------
set at 02:19:11
set at 02:19:11
Actual result:
--------------
set at 02:19:11
Fatal error: Only variables can be passed by reference in ...
Comment:
--------------
If it is passed as variable in the other example, why isn't it passed
as variable in this one? Maybe php is bogus, but I don't think this
bug-report is ...
------------------------------------------------------------------------
[2005-04-21 17:18:00] [EMAIL PROTECTED]
This is how it works. You _are_ passing it a variable.
------------------------------------------------------------------------
[2005-04-21 15:46:03] php at thoftware dot de
O.k., this for a new start, sorry for inconvenience:
Description:
------------
array_pop() removes data from objects via a method.
Seems that the first call with a method isn't recognized as being not a
variable =8-0.
Reproduce code:
---------------
class foobar {
var $cache = array();
function cache($j) {
if (!isset($this->cache[$j])) {
$this->cache[$j] = array(
'wert' => $j,
'text' => 'value: '.$j.'<br>',
'time' => 'set at '.date('h:i:s').'<br>',
);
}
return($this->cache[$j]);
}
}
$foobar =& new foobar();
echo array_pop($foobar->cache(3));
echo array_pop($foobar->cache(3));
echo array_pop($foobar->cache(3));
echo array_pop($foobar->cache(3));
Expected result:
----------------
set at 02:19:11
set at 02:19:11
set at 02:19:11
set at 02:19:11
resp.
Fatal error: Only variables can be passed by reference in ...
Actual result:
--------------
set at 02:19:11
value: 3
3
------------------------------------------------------------------------
[2005-04-21 15:27:34] [EMAIL PROTECTED]
Then please explain the problem more clearly because the code you gave
is not understandable.
The main questions are:
what did you expect?
what did you get?
Also, please try newer version before reporting (the latest 4.3.x
version ATM is 4.3.11).
------------------------------------------------------------------------
The remainder of the comments for this report are too long. To view
the rest of the comments, please view the bug report online at
http://bugs.php.net/32789
--
Edit this bug report at http://bugs.php.net/?id=32789&edit=1
