Ask for explanation -- possibly repeated

From: Benoit de Boursetty (
Date: Wed Dec 08 1999 - 06:07:15 MET

Date: Wed, 8 Dec 1999 06:07:15 +0100 (MET)
From: Benoit de Boursetty <>
Subject: Ask for explanation -- possibly repeated


  My question must already have been asked, but the error I get from the
compiler does not seem to be listed in the documentation
("This kind of expression is not allowed as right-hand side of `let rec'")

  So here's the situation. I have a basic, standard structure for trees.

# type 'a classical_tree = Classical_node of 'a * 'a classical_tree list

And now I'd like to move to another tree structure (cyclic by default)
where each node or leaf points to its parent node, if there is one.

So I want to use the following type:

# type 'a tree = Node of 'a tree option * 'a * 'a tree list

which means: Node (parent node, data, sons).

I can easily create such typed values by hand:

# let rec root = Node (None, 0, [son1; son2])
  and son1 = Node (Some root, 1, [])
  and son2 = Node (Some root, 2, []);;

But I can't manage to write the translation function from the first type
to the second one in a purely applicative fashion. It is intuitive to

# let translate_tree =
    let rec aux father (Classical_node (data, sons)) =
      let rec this_node = Node (father, data,
                                  (aux (Some this_node))
      in this_node
    in aux None

but of course this doesn't work ("This kind of expression is not allowed
as right-hand side of `let rec'", as they say) and I understand why. It's
because the contents of "this_node", not yet defined, could be looked at
in function "aux". In fact this is a correct algorithm only because
argument "father" is not accessed to in function "aux".

So, is there an applicative workaround? I know how to do it with mutable
values / references, but...

Benoit de Boursetty.

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