ID: 32789
User updated by: php at thoftware dot de
Reported By: php at thoftware dot de
Status: Open
Bug Type: Arrays related
Operating System: *
PHP Version: 4.3.11
New Comment:
And again using an object:
Reproduce code:
---------------
class foobar {
var $cache = array();
function cache() {
if (!isset($this->cache)) {
$this->cache = array(
'this is my cache<br>',
'it\'s mine<br>',
'noone else should be able to change it<br>',
'it\'s forever mine<br>',
);
}
return($this->cache);
}
}
$foobar =& new foobar();
echo array_pop($foobar->cache());
echo array_pop($foobar->cache());
echo array_pop($foobar->cache());
echo array_pop($foobar->cache());
Expected result:
----------------
it's forever mine
it's forever mine
it's forever mine
it's forever mine
Actual result:
--------------
Fatal error: Only variables can be passed by reference in ...
Using a static variable within the method will work like the example
using a plain function, using an object-variable causes a fatal error.
Previous Comments:
------------------------------------------------------------------------
[2005-04-22 15:38:05] php at thoftware dot de
How about this one?
Reproduce code:
---------------
function cache() {
static $cache=array();
if (!count($cache)) {
$cache[1] = array(
'this is my cache<br>',
'it\'s mine<br>',
'noone else should be able to change it<br>',
'it\'s forever mine<br>',
);
}
return($cache[1]);
}
echo array_pop(cache());
echo array_pop(cache());
echo array_pop(cache());
echo array_pop(cache());
Expected result:
----------------
it's forever mine
it's forever mine
it's forever mine
it's forever mine
Actual result:
--------------
it's forever mine
noone else should be able to change it
it's mine
this is my cache
Note: If you remove the '[1]'-part, it works as expected. I think this
part of the problem is not an array_pop()-bug but a bug within the
reference-system?
------------------------------------------------------------------------
[2005-04-22 14:31:53] php at thoftware dot de
The variables in the object must not be changed! That's the reason why
I don't use &cache()! Either should PHP state that it's not allowed to
use cache() as an argument for array_pop() or array_pop() shouldn't
change the data of the object because cache isn't supposed to return a
reference.
Look at the first example - expected result is an unchanged object
resp. an error-message. The second example was only to show to you,
that PHP recognizes that the method _isn't_ passed as a variable when
it is assigned to a variable before (that's bogus in my eyes).
Why do you make this bug-report bogus? I'm sorry for my maybe bad
english, but is it really not understandable what I'm trying to
explain? Then you may simply take the fact, that
Reproduce code:
---------------
$foobar =& new foobar();
echo array_pop($foobar->cache(3));
$v = $foobar->cache(3);
echo $v['time'];
works in some way, while
Reproduce code:
---------------
$foobar =& new foobar();
$v = $foobar->cache(3);
echo $v['time'];
echo array_pop($foobar->cache(3));
causes an fatal error and accept this as a bug?
------------------------------------------------------------------------
[2005-04-22 00:04:43] [EMAIL PROTECTED]
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]
------------------------------------------------------------------------
[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 ...
------------------------------------------------------------------------
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
