On 13 January 2004 15:10, Sebastian Ossio wrote: > Hi, > > I´m trying to give an empty reference as a parameter, in a > function. Or > rather make a parameter that is a reference optional. > > I have a function that should react differently if it is > given an object > or not. It works, but it gives me a warning that the second > argument is > not being given obviously. > > Sadly I cannot make the parameter optional in the function definition > because mine has to be a reference. One can´t do such a thing as > > function( $parameter1, & objectparameter = NULL) > > this results into a parsing error. > > I cannot believe that there is no way of making an object parameter > optional. > > So tell me please what I´m missing.
PHP5 ;). This is one of the improvements that will be in PHP 5 when it is released. Cheers! Mike --------------------------------------------------------------------- Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning & Information Services, JG125, James Graham Building, Leeds Metropolitan University, Beckett Park, LEEDS, LS6 3QS, United Kingdom Email: [EMAIL PROTECTED] Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
