Is it possible to deconstruct an optional argument as you can with a labeled 
argument:

  let f ~p:(x,y) () = x - y

with something like:

  let f ?(p=0,0):(x,y) () = x - y

-- 
Dr Jon D Harrop, Flying Frog Consultancy Ltd.
http://www.ffconsultancy.com/products/?e