Here is some code: $a = new my_object; $b = $a; My understanding of this operation under PHP 5+ is that $b will now be essentially a "reference" to $a, *not* a *copy* of the $a object. Is this correct? There are cases where I strictly want a *copy* of $a stored in $b. In cases like this, I supply $a's class with a copy() method, and call it like this: $b = $a->copy(); Is this reasonable, or do people have a better/more correct way to do this? Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
