> The constructor for Content accepts an > object ... why is &$db used instead of $db?
&$variable means to pass the variable by reference (as opposed to passing by value, which is the default). Instead of making a copy of the variable, you are telling $this->db to point to the same point in memory as the original $db - in effect you are using two names to point to the same data. If you modify $this->db, $db will also be modified and vice versa. Only the author of the code could tell you exactly why they chose to do things this way, but one possible reason is that when you close the database link (using $this->db->disconnect() if it's PEAR::DB) it will disconnect all the pointers to that link rather than just the one local to the class (which would be the case if it was passed in by value). Passing a reference rather than a copy also means you're not storing multiple database links in memory which could affect performance if the code was called often (e.g. on a busy web site). Hope this helps, Paul -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
