Specify it only at function definition:
myFunction( &$var1, &$var2 ) {
    // ...
}
$var = myFunction( $var1, $var2 ) ;

Ignatius
_________________________
----- Original Message ----- 
From: "Jeremy Shovan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, October 09, 2003 9:54 PM
Subject: [PHP-DB] pass by reference


> How do you pass by reference? I tried
> $var = myFunction(&$var1, &$var2);
> but it gave me a warning saying that pass by reference is depreciated
> 

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