Hi ... for newbees here, I just want to mention that reference and scalar variable aren't the same. A reference is a pointer to a scalar, just like in C or C++.
<? php $a = 2;
$ref = & $a;
echo "$a <br> $ref";
?>
this should print out: "2 <br> 2".
Scalar class also exists. Look below:
<? php
class Object_t {
var $a;
function Object_t () {
$this->a = 1;
}
}
$a = new Object_t; $ref_a = &a;
echo "$a->a <br> $ref->a";
?>
again, this should echo: "1 <br> 1";
Here is another method isued in OOP to acheive on working only over reference to scalar object. Using this, you won't ever have to ask yourself if you work on a copy of the scalar or its reference. You will only possess reference to the scalar object. If you want to duplicate the scalar object, you will have to create a function for that purpose that would read by the reference the values and assign them to another scalar of the same type... or an other type, it is as you wish at that moment.
<?php
class objet_t {
var $a;
function object_t
{
$this->a = "patate_poil";
}
}
function &get_ref($object_type)
{
return &new $object_type;
}
$ref_object_t = get_ref(object_t);
echo "$ref_object_t->a <br>";
?>
this should echo: "patate_poit <br>".
The only thing that I try to demonstrate is that scalar variable ARE object in memory while a reference is usualy a variable (scalar object) that contain the address of another scalar object, which contain the informations you want by using the reference.
Good Luck!
otek is popanowel HAT hotmailZ DOT cum
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