Hi,
Thank you all for your help! I think we're almost there: with Wolfgang's
suggestion, the Node module complies with the original constraints, as the
code below shows: (the compiler complains -- as it should! -- when on the
last line I try to nest two link nodes directly)
module rec Node:
sig
type nonlink_node_t = [ `Text of string | `Bold of Node.super_node_t list ]
type link_node_t = [ `See of string | `Mref of string * nonlink_node_t list
]
type super_node_t = [ nonlink_node_t | link_node_t ]
val text: string -> nonlink_node_t
val bold: [< super_node_t] list -> nonlink_node_t
val see: string -> link_node_t
val mref: string -> nonlink_node_t list -> link_node_t
end =
struct
type nonlink_node_t = [ `Text of string | `Bold of Node.super_node_t list ]
type link_node_t = [ `See of string | `Mref of string * nonlink_node_t list
]
type super_node_t = [ nonlink_node_t | link_node_t ]
let text txt = `Text txt
let bold seq = `Bold (seq :> super_node_t list)
let see ref = `See ref
let mref ref seq = `Mref (ref, seq)
end
open Node
let foo1 = text "foo" (* valid *)
let foo2 = bold [text "foo"] (* valid *)
let foo3 = mref "ref" [text "foo"] (* valid *)
let foo4 = mref "ref" [see "ref"] (* invalid *)
Now, the icing on the cake would be the possibility to build a Node_to_Node
set of functions without code duplication. Ideally, I would like to do
something as follows:
module Node_to_Node =
struct
open Node
let rec convert_nonlink_node = function
| `Text txt -> Node.text txt
| `Bold seq -> Node.bold (List.map convert_super_node seq)
and convert_link_node = function
| `See ref -> Node.see ref
| `Mref (ref, seq) -> Node.mref ref (List.map convert_nonlink_node seq)
and convert_super_node node = match node with
| #nonlink_node_t -> (convert_nonlink_node node :> super_node_t)
| #link_node_t -> (convert_link_node node :> super_node_t)
end
But this fails on the last line:
Error: This expression has type
[< `Bold of 'a list & Node.super_node_t list | `Text of string ]
as 'a
but is here used with type
[< `Mref of string * 'a list | `See of string ]
These two variant types have no intersection
Do you see any way around it?
Thanks again for all your time and attention!
Cheers,
Dario
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