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To: Caml List <caml-list@inria.fr>
From: Joel Reymont <joelr1@gmail.com>
Subject: camlp4 3.10: Hashtbl.t and <:ctyp< $tp1$ . $tp2$ >>
Date: Mon, 30 Apr 2007 22:52:50 +0100
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X-Spam: no; 0.00; camlp:01 hashtbl:01 ctyp:01 hashtbl:01 denoting:01 ctyp:01 camlp:01 rec:01 rec:01 assert:01 assert:01 strings:01 ident:02 ident:02 tree:02 

Consider the following function. My understanding is that for, say,  
Hashtbl.t it should return  a list of types where the 1st element  
represents "t" and the second represents "Hashtbl".

   (* [tp_path tp] takes a type. @return a module path (list of strings)
      denoting this type. *)
   let rec tp_path = function
     | <:ctyp< $lid:id$ >> | <:ctyp< $uid:id$ >> -> [id]
     | <:ctyp< $tp1$ . $tp2$ >> ->
        (match tp_path tp2 with [n] -> n | _ -> assert false) ::  
tp_path tp1
     | _ -> invalid_arg "tp_path"

Nicolas suggested the following approach which doesn't work because  
tp1 and tp2 become idents. There's a function to convert from ctyp to  
ident but I could not find the opposite from searching the camlp4  
source tree.

   let rec tp_path = function
     | <:ctyp< $lid:id$ >> | <:ctyp< $uid:id$ >> -> [id]
     | <:ctyp< $id:id$ >> ->
       (match id with
          | <:ident< $tp1$ . $tp2$ >> ->
            (match tp_path tp2 with [n] -> n | _ -> assert false) ::  
tp_path tp1
          | _ -> invalid_arg "tp_path")
     | _ -> invalid_arg "tp_path"

Since Hashtbl.t is a valid type, there should be a camlp4 ctyp  
pattern to match it, no?

Am I missing something obvious?

	Thanks, Joel

--
http://wagerlabs.com/