[Haskell-cafe] function composition and Rank2Types

[Brandon Allbery]

Yes; a more accurate statement to what I said earlier is that `($)` is
special-cased to always behave as if `ImpredicativeTypes` is on,
whereas `(.)` requires you to enable it explicitly.

On Fri, Feb 3, 2023 at 1:32 PM Hage, J. (Jurriaan) via Haskell-Cafe
<haskell-cafe at haskell.org> wrote:
>
> If you turn on ImpredicativeTypes it does compile.
>
> Jur
>
>
> > On 3 Feb 2023, at 16:55, Olaf Klinke <olf at aatal-apotheke.de> wrote:
> >
> > Dear Cafe,
> >
> > I used to believe that
> >   f . g $ x
> > is equivalent to
> >   f $ g x
> > but apparently that is not the case when rank-2 types are involved.
> > Counterexample:
> >
> > {-# LANGUAGE RankNTypes #-}
> > type Cont = forall r. (() -> r) -> r
> > data Foo = Foo Cont
> >
> > mkCont :: () -> Cont
> > mkCont x = \f -> f x
> >
> > foo1 :: () -> Foo
> > foo1 x = Foo $ mkCont x
> >
> > foo2 :: () -> Foo
> > foo2 x = Foo . mkCont $ x
> >
> > The function foo1 type-checks, the function foo2 doesn't. And that is
> > although ghci tells me that
> > mkCont :: () -> Cont
> > Foo    :: Cont -> Foo
> > so I should be able to compose? What is going on here? GHC 9.0.2 errors
> > with
> >    • Couldn't match type ‘b0’ with ‘Cont’
> >      Expected: b0 -> Foo
> >        Actual: Cont -> Foo
> >      Cannot instantiate unification variable ‘b0’
> >      with a type involving polytypes: Cont
> >    • In the first argument of ‘(.)’, namely ‘Foo’
> >      In the expression: Foo . mkCont
> > and GHC 8.8.4 errors with
> >    • Couldn't match type ‘(() -> r0) -> r0’ with ‘Cont’
> >      Expected type: ((() -> r0) -> r0) -> Foo
> >        Actual type: Cont -> Foo
> >    • In the first argument of ‘(.)’, namely ‘Foo’
> >      In the expression: Foo . mkCont
> >
> > The latter suggests that r is unnecessarily specialized to some
> > specific-yet-unknown r0 which destroys the universal quantification.
> >
> > Olaf
> >
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-- 
brandon s allbery kf8nh
allbery.b at gmail.com