[Haskell-cafe] function composition and Rank2Types

[Olaf Klinke]

On Fri, 2023-02-03 at 18:31 +0000, Hage, J. (Jurriaan) wrote:
> If you turn on ImpredicativeTypes it does compile.
> 
> Jur

Not under GHC 8.8.4. 
In any case I find it rather disturbing that Hask is not a category!
That is, there are types a, b, c and morphisms f :: a -> b, g :: b -> c
such that g.f is not a morphism. The current base-4.17 instance
Category (->) uses (GHC.Base..) as (.) so it is a lie, since the
Control.Category module does not use ImpredicativeTypes. It should
read 

instance Category (->) where
  id = GHC.Base.id
  g.f = \x -> g $ f x

Indeed, the recent ($) has a type much funnier than (.). Did any of the
teaching members of the Haskell Café have difficulties with this? I had
the vague feeling of deja-vu (all this been discussed) when writing the
initial message.  

btw, the link in https://wiki.haskell.org/Impredicative_types to the
GHC user guide is a 404. The correct current link is
https://downloads.haskell.org/ghc/latest/docs/users_guide/exts/impredicative_types.html
Maybe someone with a wiki account can change that. Only on the latter
page one can find Brandon's explanation about the special behaviour of
($) and that while older GHC versions did have the ImpredicativeTypes
extension, it was not reliable, as I have witnessed. There is also no
mention of ImpredicativeTypes in any of the error messages I posted,
but I think there is already an open GHC issue for that:
https://github.com/haskell/error-messages/issues/521
Should this go into the Haskell error message index until fixed? 

Apologies for being so pedantic but this shook my belief in Haskell as
a model of mathematical functions. 

Olaf



> 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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