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Date: Fri, 30 Aug 2002 10:21:18 +0200
From: Daniel de Rauglaudre <daniel.de_rauglaudre@inria.fr>
To: caml-list@inria.fr
Subject: Re: [Caml-list] camlp4: pretty printing not to a file
Message-ID: <20020830102118.B8876@verdot.inria.fr>
References: <5B049D08-BB80-11D6-9EB3-00039375801A@yahoo.com>
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In-Reply-To: <5B049D08-BB80-11D6-9EB3-00039375801A@yahoo.com>; from jehenrik@yahoo.com on Thu, Aug 29, 2002 at 02:51:41PM -0400
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Hi,

On Thu, Aug 29, 2002 at 02:51:41PM -0400, jehenrik wrote:

> Also it seems like there might be a way to take the parsing offsets
> and look back in the original pre-parsed string to get the real
> source snippet.

   let getsource fname (bp, ep) =
     let ic = open_in_bin fname in
     let s = String.create (ep - bp) in
     seek_in ic bp;
     really_input ic s 0 (ep - bp);
     close_in ic;
     s

> EXTEND
>    expr: LEVEL "expr1"
>    [ LEFTA [ "REFLECTLIST"; hd = expr; "::"; tl = expr ->
>      map_expr reflect_arg <:expr< [$hd$::$tl$] >>
>    ]  ];
> END;;
> 
> let r = Grammar.Entry.parse expr (Stream.of_string "REFLECTLIST 
> [1;3;5;11]");;

Input: REFLECTLIST [1;3;5;11]
The rule recognizes the keyword REFLECTLIST.
Then the rule symbol "hd = expr" recognizes the expr "[1;3;5;11]"
Then the rule expects a "::".
There is no "::".
Therefore: syntax error.

> let r = Grammar.Entry.parse expr (Stream.of_string "REFLECTLIST 
> 1::3::5::11::[]\
> ");;

Idem. The symbol "hd = expr" recognizes the expr "1::3::5::11::[]".
There is no "::" following it.
Therefore syntax error again.

> Can anyone please enlighten me?

And why not just:

EXTEND
   expr: LEVEL "expr1"
   [ LEFTA [ "REFLECTLIST"; e = expr ->
     map_expr reflect_arg e
   ]  ];
END;;

-- 
Daniel de RAUGLAUDRE
daniel.de_rauglaudre@inria.fr
http://cristal.inria.fr/~ddr/
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