ID: 44347 Updated by: [EMAIL PROTECTED] Reported By: rewilliams at newtekit dot com -Status: Open +Status: Bogus Bug Type: Class/Object related Operating System: OS X 10.2.5 PHP Version: 5.2.5 New Comment:
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 See http://docs.php.net/manual/en/language.oop5.basic.php Previous Comments: ------------------------------------------------------------------------ [2008-03-05 23:35:21] rewilliams at newtekit dot com Description: ------------ If you try to use this syntax in a class to declare a class member: $foo[1] = 1; You get a parse error. Prepending a visibility keyword doesn't change anything, nor does using a string key instead of a numeric key. Note: this was seen in v5.2.4. I don't have 5.2.5 to test against. I did, however, test it against a couple of 5.0.x versions, where I also saw a parse error, so this is definitely a multi-version issue. Reproduce code: --------------- File test.php <?php class foo { $bar[1] = 1; } ?> Expected result: ---------------- I'd expect $bar[1] to be created as member of class foo and assigned the value of 1. This syntax isn't explicitly covered in the classes section of the docs as far as I can see; the only restriction that's given is this: "The default value must be a constant expression, not (for example) a variable, a class member or a function call." Since I'm assigning a constant value, since this array assignment syntax is otherwise valid, and since the array() syntax works, I'd expect this syntax to work, as well. Actual result: -------------- % php -l test.php Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION in test.php on line 4 Errors parsing test.php ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=44347&edit=1
