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Subject: Re: [Caml-list] How to achieve this camlp4 syntax extension
From: David Teller <David.Teller@univ-orleans.fr>
To: Conglun Yao <yaoconglun@gmail.com>
Cc: caml-list@yquem.inria.fr
In-Reply-To: <f7b50d2a0904020442rb51fe54p496e6f2f2b665c3e@mail.gmail.com>
References: <f7b50d2a0904020442rb51fe54p496e6f2f2b665c3e@mail.gmail.com>
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Date: Thu, 02 Apr 2009 14:40:07 +0200
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X-Spam: no; 0.00; camlp:01 syntax:01 univ-orleans:01 camlp:01 frisch:01 cheers:01 0100,:01 syntax:01 expr:01 expr:01 longident:01 'list':01 longident:01 beginner's:01 ocaml:01 

   Hi,
 I'm not going to quite answer your question yet -- I'd need to check
the source of Camlp4 for this. However, I can point out that what you're
trying to do looks very much like pa_open, by Alain Frisch.

let _ = open M1.M2 in e

will execute [e] using module [M1.M2].

Cheers,
 David

On Thu, 2009-04-02 at 12:42 +0100, Conglun Yao wrote:
> Dear all,
> 
> I tried to achieve the following syntax extension, but failed.
> 
> Add expression .[ ] after a module name, inside .[ ] I want to refer
> to the specified module, like
> 
> let _ = M1.M2.[ here is my syntax,  using M1.M2 module ]
> 
> Here is my attempt (failed)
> 
> EXTEND Gram
> 
> GLOBAL: expr;
> 
> expr: LEVEL "top"[
> [ e1 = module_longident; ".";  "[";  (*t = test_syntax;*)  "]" ->
>   <:expr< $id:e1$ test >>
> ]];
> 
> END
> 
> Different kinds of error happened, when trying to use it.
> 
> Even the ordinary expression:  List.length [1; 2;3 ],  failed.  'List'
> is parsed as module_longident, try to match the rule I defined.
> 
> Thanks for any help.
> 
> Conglun
> 
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