Pointer in PHP

[Denis Crespe]

Hi,

Is it possible to keep a pointer to an object in a PHP variable? I'd like to
do this :

$obj1 = new myClass(...);
$obj2 = $obj1;
$obj1->name = "foo";
$obj2->name = "bar";

echo $obj1->name;

Which would display "bar" instead of "foo" with the current PHP behavior.

Thanks a lot,
Denis

[rush]

"Denis Crespe" <[e-mail]> wrote in message
news:bn******** **@news.tiscali .fr...

$obj1 = new myClass(...);
$obj2 = $obj1;
$obj1->name = "foo";
$obj2->name = "bar";

echo $obj1->name;

Which would display "bar" instead of "foo" with the current PHP behavior.

you can not get the pointer, but you can get the reference:

$obj2 = & $obj1;

rush
--
http://www.templatetamer.com/

[Denis Crespe]

Awesome! It's what I wanted :-)

Denis

"rush" <pi**@rush.aval on.hr> a écrit dans le message de news:
bn**********@ls 219.htnet.hr...

"Denis Crespe" <[e-mail]> wrote in message
news:bn******** **@news.tiscali .fr...

$obj1 = new myClass(...);
$obj2 = $obj1;
$obj1->name = "foo";
$obj2->name = "bar";

echo $obj1->name;

Which would display "bar" instead of "foo" with the current PHP

behavior.
you can not get the pointer, but you can get the reference:

$obj2 = & $obj1;

rush
--
http://www.templatetamer.com/

[John Downey]

Denis Crespe wrote:

Hi,

Is it possible to keep a pointer to an object in a PHP variable? I'd like to
do this :

$obj1 = new myClass(...);
$obj2 = $obj1;
$obj1->name = "foo";
$obj2->name = "bar";

echo $obj1->name;

Which would display "bar" instead of "foo" with the current PHP behavior.

Thanks a lot,
Denis

$obj2 = &$obj1;

--
John Downey
http://delusive.dyn.ee
http://sage.dev.box.sk
http://blacksun.box.sk