Deconstructing optional arguments

Deconstructing optional arguments

[Jon Harrop]

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

Re: [Caml-list] Deconstructing optional arguments

[Alain Frisch]

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

Re: [Caml-list] Deconstructing optional arguments

[Olivier Andrieu]

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On Fri, Apr 11, 2008 at 11:24 AM, Jon Harrop <jon@ffconsultancy.com> 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

sure, cf. the grammar in the refman
http://caml.inria.fr/pub/docs/manual-ocaml/expr.html

parameter	::=
        ...
 	¨O	 ? label-name : (  pattern  [: typexpr]  [= expr] )


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

-- 
  Olivier