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[Caml-list] First order compile time functorial polymorphism in Ocaml
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 Date: -- (:) From: Michal Moskal Subject: Re: [Caml-list] First order compile time functorial polymorphism in Ocaml
```On Mon, Jun 23, 2003 at 01:52:14PM +1000, John Max Skaller wrote:
> Michal Moskal wrote:
>
> >On Mon, Jun 23, 2003 at 04:25:18AM +1000, John Skaller wrote:
>
> >First thing to consider: map function of this kind only exists for types,
> >where type variables occur only positively.
>
> What does that mean?

Variable occurs positively in type, if it's preceded by even number of
negations.

When you treat -> as logical implication, you get:

a -> b == a & !b

So 'a -> t, 'a -> (t -> 'a) are 'a-positive, t -> 'a is 'a-negative,
and 'a -> 'a isn't neither 'a positive nor 'a negative. Other tycons
(like *) doesn't change sign. But when you define new type, say:

type ('a, 'b) t = Foo of 'a -> 'b

then ('c, 'd) t is 'd negative, and 'c positive.

> >Even then, for inductive
> >types it requires proof (it isn't as simple as it first seems).
>
>
> The proof has been constructed for all polynomial types,
>
> i.e. types using only sums and products.
> [Paper:Functorial ML, Author:Barry Jaye, the mechanism
> there generalises over 'arbitrary' algorithms: I'm not proposing
> that, rather that the theory can be applied to hand write
> the generators for popular functions like map and fold]

Ah, you mean only sums and products. Then you're correct (you are
avoiding the hard case :-).

> >For example consider:
> >
> >type 'a t = Foo 'a -> unit
> >
> >To map : 'a t -> 'b t here, you need f : 'b -> 'a.
>
>
> Ah, ok, exponential is contravariant.
>
> Probably have to think about
>
> 	'a ref
>
> too.

type 'a ref = {mutable contents : 'a}

which is much the same as product type.

--
: Michal Moskal :: http://www.kernel.pl/~malekith : GCS {C,UL}++++\$ a? !tv
: When in doubt, use brute force. -- Ken Thompson : {E-,w}-- {b++,e}>+++ h

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