#32789 [Opn]: array_pop doesn't work correctly

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: Sorry, the fatal error results from the result of cache() being an empty array (but that may be another bug?), correct code should be: Reproduce code: --------------- class foobar { var $cache = array(); function cache() { if (!count($this->cache)) { $this->cache[1] = array( 'this is my cache<br>', 'it\'s mine<br>', 'noone else should be able to change it<br>', 'it\'s forever mine<br>', ); } return($this->cache[1]); } } $foobar =& new foobar(); echo array_pop($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 will work like shown above, even if you remove the '[1]'-part. Previous Comments: ------------------------------------------------------------------------ [2005-04-22 15:46:13] php at thoftware dot de 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. ------------------------------------------------------------------------ [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] sniper@php.net 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 php-general@lists.php.net ------------------------------------------------------------------------ [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). ------------------------------------------------------------------------ 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