Date: Wed, 8 Dec 1999 06:07:15 +0100 (MET)
From: Benoit de Boursetty <email@example.com>
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 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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