Re: Empty reference parameter

Hi, Wednesday, January 14, 2004, 1:10:53 AM, you wrote: SO> Hi, SO> I´m trying to give an empty reference as a parameter of a function. Or SO> rather, make a parameter that is a reference optional. SO> I have a function that should react differently if it is given an object SO> or not. It works, but it gives me a warning that the second argument is SO> not being given obviously. SO> Sadly I cannot make the parameter optional in the function definition SO> because mine has to be a reference. One can´t do such a thing as SO> function( $parameter1, & objectparameter = NULL) SO> this results into a parsing error. SO> I cannot believe that there is no way of making an object parameter SO> optional. SO> So tell me please what I´m missing. Try doing it in 2 steps $ref = false; function test(&$var) if(is_object($var)){ echo "object passed"; } } test($ref); $ref =& new object; test($ref); -- regards, Tom