NYCPHP Meetup

[Tim Gales]

Is there some reason this 
will not work in your situation:

function tst_func ($statement, &$test1) {
	if ( !isset ($test1) ) {
		$test1 = "the usual";
	}
	echo $statement . $test1 . "\n";
}

$tst_str  = "something new";
$tst_ref = & $tst_str;

tst_func( "Stuff was: ",  $tst_ref );
tst_func( "Stuff was: " );


T. Gales & Associates
'Helping People Connect with Technology'

http://www.tgaconnect.com

p.s. kinda close to what was already posted
     sorry for the redundancy and repeating


> -----Original Message-----
> From: talk-bounces at lists.nyphp.org
> [mailto:talk-bounces at lists.nyphp.org] On Behalf Of Daniel Convissor
> Sent: Friday, January 16, 2004 11:34 AM
> To: NYPHP Talk
> Subject: [nycphp-talk] passing optional arguments by reference
> 
> 
> 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?
> 
> Thanks,
> 
> --Dan
> 
> -- 
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