NYCPHP Meetup

[Dan Cech]

Daniel Convissor wrote:

> Hi Folks:
> 
> I'm want the parameters for a function to be passed by reference.  
> Normally, that's accomplished by placing a & in front of the variable. 
> Trick is, the argument in question needs to be optional.  Placing a & in 
> front of an optional parameter creates a parse error.
> 
> CODE:
>     function &execute($stmt, &$data = array()) {
>     }
> 
> ERROR:
>     Parse error: parse error, unexpected '=', expecting ')'
> 
> Removing the "= array()" makes it parse fine.
> 
> So, is there a way to do this, please?

AFAIK, no.

There is a workaround though.  If you create an array containing a 
reference, then pass that array normally, your function will work on a 
copy of the array, which contains a copy of the reference, which will 
still reference the same object.

If that didn't make any sense:

$myobj =& new stdClass ();

print_r ($myobj);

$myarr = array ();
$myarr['myobj'] =& $myobj;

function myfunc ($input = array ())
{
	if ( isset ($input['myobj']) )
	{
		$input['myobj']->test = TRUE;
	}
}

myfunc ($myarr);

print_r ($myobj);

Hope this helps,

Dan

> Thanks,
> 
> --Dan
>