Hi, Reading the PHP 5 documentation at: HYPERLINK "http://www.php.net/manual/en/language.oop5.basic.php"http://www.php.net/man ual/en/language.oop5.basic.php, I am confused. In the example given, what is the difference between: $assigned = $instance; $reference =& $instance; I would expect all of the var_dump to display NULL The doc says "When assigning an already created instance of an object to a new variable, the new variable will access the same instance as the object that was assigned." so the above assignments seem the same to me and setting $instance to NULL should also set $assigned to NULL. If this is not the case and not using the '&' specifies a 'copy' (contradicting the documentation) then what's the purpose of object cloning? I tried the code below and find that it gives the exact same output regardless if I am using the '&' or not so it seems to assign be reference either way. <?php class SimpleClass { // member declaration public $var = 'a default value'; // method declaration public function displayVar() { echo $this->var; } } $instance = new SimpleClass(); $assigned = $instance; // $assigned = &$instance; // No difference if this line is used instead $instance->var = 'Value has been changed'; var_dump($instance); echo '<br /><br />'; var_dump($assigned); ?> -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.300 / Virus Database: 265.6.8 - Release Date: 1/3/2005 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
