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From: Michael Hicks <mwh@cs.umd.edu>
To: "'Oleg'" <oleg_inconnu@myrealbox.com>, caml-list@inria.fr
Subject: RE: [Caml-list] Cyclic ?!
Date: Thu, 15 Aug 2002 10:31:19 -0400	[thread overview]
Message-ID: <000301c24468$6cb5b350$6601a8c0@hicks> (raw)
In-Reply-To: <200208150218.WAA22018@hickory.cc.columbia.edu>

If memory serves, ML datatypes elegantly overload about three distrinct
type-theoretic constructs.  In particular, the variant tags in the data
type definition double as the "witnesses" to coercions between an
iso-recursive type and its "unrolling" of one level.  Without going into
detail, this is why you can define something like

type foo = Rec of foo;;

but you can't define

type foo = foo;;

The latter definition is in the style of "equi-recursive" types, in
which a recursive type is equivalent any number of its unrollings.  This
formulation is harder to typecheck, as I recall.  There is a similar
analogy between the two definitions of bin_tree that you present.

Mike

-----Original Message-----
From: owner-caml-list@pauillac.inria.fr
[mailto:owner-caml-list@pauillac.inria.fr] On Behalf Of Oleg
Sent: Wednesday, August 14, 2002 10:20 PM
To: caml-list@inria.fr
Subject: [Caml-list] Cyclic ?!

Hi

I'm puzzled by the following compiler behavior:

If I define bin_tree as

type 'a bin_tree = 
           Empty
         | Node of 'a bin_tree * 'a * 'a bin_tree

the compiler accepts it. OTOH if I define it as

type 'a bin_tree = ('a bin_tree * 'a * 'a bin_tree) option

It gives an error: "The type abbreviation bin_tree is cyclic".
Why??? And what's the difference between the two, really?

Thanks
Oleg
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  reply	other threads:[~2002-08-17 10:59 UTC|newest]

Thread overview: 7+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2002-08-15  2:19 Oleg
2002-08-15 14:31 ` Michael Hicks [this message]
2002-08-15 17:26   ` Oleg
2002-08-15 18:05     ` Markus Mottl
2002-08-15 18:16       ` Brian Smith
2002-09-24 16:23     ` [Caml-list] Recursive types (Was Cyclic ?!) Christophe Raffalli
2002-08-18 16:13 ` [Caml-list] Cyclic ?! John Max Skaller

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