ID: 11909
Updated by: brianlmoon
Reported By: [EMAIL PROTECTED]
Old-Status: Open
Status: Bogus
Bug Type: Scripting Engine problem
Operating system:
PHP Version: 4.0.6
Assigned To:
Comments:
This is expected behavior. You are assuming that the object you had in $obj2 should
be gone when you copy $obj to that var. That is not the case. It only changes what
$obj2 points to. Take this case:
$var=3;
$var2=&$var;
unset($var);
echo $var2;
$var2 is still 3. The reference is gone but not the value of the var. That is what
makes references nice.
Brian.
Previous Comments:
---------------------------------------------------------------------------
[2001-07-05 10:54:01] [EMAIL PROTECTED]
When objects that contain objects that are referenced
elsewhere, they are not correctly dereferenced, and strange
side effects result
class Foo{
var $a;
function Foo($value)
{
$this->set($value);
}
function set($value)
{
$this->a = $value;
}
function get()
{
return $this->a;
}
}
class Bar{
var $a;
function Bar($value)
{
$this->a = new Foo($value);
}
function set($value)
{
$this->a->set($value);
}
function get()
{
return $this->a->get();
}
function share($other)
{
$this->a = & $other->a;
}
}
//Objects are initialized
$obj = new Bar(1);
$obj2 = new Bar(2);
//$obj and $obj2 both contain an internal reference to the
//same Foo object, with a value of 2
$obj->share($obj2);
//$obj2 now references a new Bar object with a new
//Foo object containing a value of 3
$obj2 = new Bar(3);
//$obj2 should be a copy of $obj1 and contain a new
//Foo object with a value of 2
$obj2 = $obj;
//Somehow, this also changes the value in $obj2
$obj->set(4);
//In a sane world, this outputs 2
//In reality, we get a 4
echo $obj2->get();
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