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.
Correct. Given only these assignments, $instance is NULL and so $assigned and $reference would also be 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?
This was pretty much only needed in PHP4. It might help you to read the PHP4 OOP documentation notes. In PHP5 you don't need the & to signify an object as a reference; references are passed by default instead of copies.
http://php.net/manual/en/language.oop.php
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
In PHP5, the above comment is true. In PHP4, it is not true.
$instance->var = 'Value has been changed';
var_dump($instance);
PHP4: 'Value has been changed' PHP5: 'Value has been changed'
echo '<br /><br />'; var_dump($assigned);
PHP4: 'a default value' (if you use the &, this would read 'Value has been changed')
PHP5: 'Value has been changed'
?>
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