ID: 36930 User updated by: ante at abstraktmedia dot com Reported By: ante at abstraktmedia dot com Status: Bogus Bug Type: SPL related Operating System: Win XP PHP Version: 5.1.2 New Comment:
As you wish Mike but this is just very very bad workaround....this checking should be done internaly not on user end... Previous Comments: ------------------------------------------------------------------------ [2006-03-31 11:56:21] [EMAIL PROTECTED] Thank you for taking the time to write to us, but this is not a bug. Please double-check the documentation available at http://www.php.net/manual/ and the instructions on how to report a bug at http://bugs.php.net/how-to-report.php if (($foo instanceof ArrayObject) || is_array($foo)) ------------------------------------------------------------------------ [2006-03-31 11:46:51] ante at abstraktmedia dot com Description: ------------ OK... I'm using ArrayObject quite a lot and I experienced something which is not logical... If I have $arrayObj = new ArrayObject(Array("Test")) function is_array() returns false which is not ok in the first place...but never mind...we'll cast it to array so we use if(is_array((array) $arrayObj)) ..... but the problem is what if $arrayObj is not Array...what if it's null or string? casting to array will make the function return true even if the variable is null.... That's the real problem.... Shor story...I cannot use is_array on ArrayObject if I don't cast that object to a "normal" array..But if I use casting and variable is for example string then is_array will return true when it should return false.... Reproduce code: --------------- Logicaly: $arrayObj = new ArrayObject(Array("Test")); //we don't use casting!! if(is_array($arrayObj)) ... should return TRUE but because of the internal workings of PHP we have to convert ArrayObject to Array before sending it to is_array function... But what if our Array is String actually...??or Null??? Because we HAVE TO cast that ArrayObject before sending it to is_array it will also cast String or anything else to an Array and therefore function will return TRUE... $arrayObj = new ArrayObject(Array("Test")); //$arrayObj = null; if(is_array((array) $arrayObj)) { print("Is array!"); } else { print("Not array!"); } Now try uncommeting second line and commenting first... ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=36930&edit=1
