Jon Harrop wrote:
> 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

Yes, of course. The default value is attached to the pattern, not to the 
label:

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

-- Alain