[Haskell-cafe] Infinite types should be optionally allowed

Luke Palmer lrpalmer at gmail.com
Fri Feb 13 18:09:47 EST 2009

On Fri, Feb 13, 2009 at 4:04 PM, Luke Palmer <lrpalmer at gmail.com> wrote:

> On Fri, Feb 13, 2009 at 3:13 PM, Job Vranish <jvranish at gmail.com> wrote:
>> There are good reasons against allowing infinite types by default
>> (mostly, that a lot of things type check that are normally not what we
>> want). An old haskell cafe conversation on the topic is here:
>> http://www.nabble.com/There%27s-nothing-wrong-with-infinite-types!-td7713737.html<http://www.nabble.com/There%27s-nothing-wrong-with-infinite-types%21-td7713737.html>
>> However, I think infinite types should be allowed, but only with an
>> explicit type signature. In other words, don't allow infinite types to
>> be inferred, but if they are specified, let them pass. I think it
>> would be very hard to shoot yourself in the foot this way.
Oops!  I'm sorry, I completely misread the proposal.  Or read it correctly,
saw an undecidability hiding in there, and got carried away.

What you are proposing is called equi-recursive types, in contrast to the
more popular iso-recursive types (which Haskell uses).  There are plentiful
undecidable problems with equi-recursive types, but there are ways to pull
it off.  The question is whether these ways play nicely with Haskell's type

But because of the fundamental computational problems associated, there
needs to be a great deal of certainty that this is even possible before
considering its language design implications.

> That inference engine seems to be a pretty little proof-of-concept, doesn't
> it?  But it is sweeping some very important stuff under the carpet.
> The proposal is to infer the type of a term,  then check it against an
> annotation.  Thus every program is well-typed, but it's the compiler's job
> to check that it has the type the user intended.  I like the idea.
> But the inference engine is only half of the story.  It does no type
> checking.  Although checking is often viewed as the easier of the two
> problems, in this case it is not.  A term has no normal form if and only if
> its type is equal to (forall a. a).  You can see the problem here.
> Luke
>> Newtype is the standard solution to situations where you really need
>> an infinite type, but in some cases this can be a big annoyance. Using
>> newtype sacrifices data type abstraction and very useful type classes
>> like Functor. You can use multiparameter type classes and functional
>> dependencies to recover some of the lost abstraction, but then type
>> checking becomes harder to reason about and the code gets way more
>> ugly (If you doubt, let me know, I have some examples). Allowing
>> infinite types would fix this.
>> I'm imagining a syntax something like this:
>> someFunctionThatCreatesInfiniteType :: a -> b | b = [(a, b)]
>> Thoughts? Opinions? Am I missing anything obvious?
>> - Job
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