http://qs321.pair.com?node_id=103889


in reply to i don't know rather take a look below

In your example, $aa would get the value of $a, and @bb would get ($b, $c). I don't know what you mean by "elements of @bb, but sorted individually", because $a isn't a member of @bb.

Could you describe what you're trying to do a little better?

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Re: Re: i don't know rather take a look below
by nehlwyn (Acolyte) on Aug 10, 2001 at 19:24 UTC
    Sure ... let's take an example where $abutton is a gtk button , $a its label , $bbutton if it exists a button from the same button group . ... #depending on the circumstances @args is either equal to #($a) or ($a,$bbutton) Make_Button($abutton,@args); ... sub Make_Button { my ($aa,@bb)=@_; $aa = new Gtk::RadioButton(here must come either $a or $a,$bbutton); } #end of script The syntax is very rigid , how do i manage to have the equivalent of $button = new Gtk::RadioButton($a, $bbutton); within the sub , and starting from @bb Sorry for the confusion , maybe it's confusing because i see a problem where there's none ( but a mere $aa = new Gtk::RadioButton(@bb) within the sub doesn't work so ... P.S.: i pretty new , can you tell me where this message is visible ? i have to pass through my home node to see it ...