Posted by Don on 01/06/05 21:19

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);
?>



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