Re: Function definition: how to make default argument an empty array?

Can't test it now, but have you tried function makeyogurt ($flavour, $type = list()) I assume you need it to be an array because you want to either walk it or, more likely, perform an in_array() test on it and you found that passing non-array variables to in_array issues an error. If initializing as an array doesn't work you can always do this: function makeyogurt ($flavour, $type = "") { while ((is_array($type)) && (list($key,$val)=each($type))) { [loop] } } or this: function makeyogurt ($flavour, $type = "") { if ((is_array($type)) && (in_array("strawberry_yoghurt",$type))) { [loop] } } or use the stone-age method: function makeyogurt ($flavour, $type = "") { if (@in_array("strawberry_yoghurt",$type)) { [loop] } } Bogdan Michael Jurgens wrote: > Hi, > > As you all may know, this is how you set an optional second argument, that > defaults to acidophilus > > function makeyogurt ($flavour, $type = "acidophilus") > { } > > I'm now looking for a way to have the second (optional) argument be an array > of strings. I can't get it to work though... > > In pseudo-code: > > function makeyogurt ($flavour, $type = 'EMPTY ARRAY') > { } > > Any help would be much appreciated, > Michael > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net