Re: [PHP4BETA] Returning objects

On Wed, 24 Nov 1999, Zeev Suraski wrote: > At 21:13 24/11/1999 , Joel Palmius wrote: > >Sorry if I missed the correct part of documentation. If this is the case, > >could someone point me to the correct URL? Otherwise: > > > >Let's say I have a linked list/hierarchy: > > > > class TreeNode > > { > > var parent; > > var children; > > var nodename; > > ... > > > > Function GetFirstChild() > > { > > return $this->children[0]; > > } > > ... > > } > > > >(children is an array of TreeNode). Now let's say I have an object > >("$mynode") of the TreeNode class and want to change the nodename of the > >first child: > > > > $child = $mynode->GetFirstChild(); > > $child->nodename = "Hello"; > > > >This works ok, only what I change is a *copy* of the real child, and not > >the child itself. When doing this afterwards: > > > > $child = $mynode->GetFirstChild(); > > echo $child->nodename; > > > >.. I get the old nodename. Now, how do I get around this? Since this is > >the syntax of copying a variable itself and not the value of the variable: > > > > $a = &$b; > > > >.. I would believe the syntax of returning a variable reference (in > >"GetFirstChild()" in the above example) would be: > > > > return &$this->children[0]; > > > >or, if this isn't possible, to try to get a reference the other way: > > > > $child = &$mynode->GetFirstChild(); > > > >Neither seems to work though (I get "parse error"). So, to sum up, my > >question is how to return an object from a method in such a way that I > >can change the parameters of the object (and not of a copy of the object) > >outside of the class. Is it even possible? > > It currently isn't. It's something that's planned for a future version of > PHP, but not 4.0 (most probably 4.1). Doesn't the current way in practise exclude even the possibility of several of the most common operations in OO? From what I've gathered, most other languages would return an object reference in the example above. Wouldn't this a better way to handle things? // Joel > > Zeev > -- > Zeev Suraski <zeev@zend.com> http://www.zend.com/ > >