OOP: variables passed by reference by default?

[neylitalo]

I was wondering if someone could help me out with this - this code and the results lead me to believe that variables are being passed by reference, but my instincts tell me that they shouldn't be.

[code]$organization = $organizationBuilder->getOrganization();
$organization2 = $organizationBuilder->getOrganization();[/code]

Here, I would assume that I'd have two independent objects, right? Read on...

[code]$organization->setName("Test Organization");

die($organization2->getName());[/code]

And now, I assign $organization a name, and get the name of $organization2. However, it spits out "Test Organization".

And if I do this:
[code]echo $organization."<br />";
echo $organization2."<br />";[/code]

I get
[quote]Object id #497
Object id #497[/quote]

So $organization2 and $organization are the same object? I think I'm missing something... can somebody explain?

[willfitch]

In PHP4, they aren't passed by reference unless you append the ampersand (&) to that particular class (i.e. $var = &new Class()).  

In PHP5, the default is to pass by reference.  Also, it depends on the returned value of the method, and whether or not you have cloned that object

[neylitalo]

Damn. I was afraid of that. Thanks.

Is there any way to force it to pass by value?

[willfitch]

Hey neylitalo,

As far as I know, you can't stop passing by reference in PHP5.  However, if you are trying to make copies of an object, simply use the "clone" keyword.

Example:

[code=php:0]
<?php
$obj1 = new Class();
$obj2 = clone $obj1;
?>
[/code]

This should help your issue.

[neylitalo]

beautiful, thank you! :)

[hvle]

I really think your problem lies inside this function:
$organizationBuilder->getOrganization();

if inside that function, you creates a new instance and return, then you should not have problem.