[Haskell-cafe] Can I get 'forall (a :: k). Typeable a => Dict (Typeable k)'?

[Adam Gundry]

Hi Tom,

I think you are looking for Type.Reflection.typeRepKind

     typeRepKind :: TypeRep (a :: k) -> TypeRep k

which combined with

     typeRep :: Typeable a => TypeRep a
     withTypeable :: forall k (a :: k) rep (r :: TYPE rep). TypeRep a -> 
(Typeable a => r) -> r

should allow you to define

     kindable :: forall (a :: k). Typeable a => Dict (Typeable k)

as your subject line asked for, albeit not the message body. :-)

Cheers,

Adam


On 23/08/2023 11:50, Tom Ellis wrote:
> Morally I think that
> 
>      class Typeable (a :: k)
> 
> should have been
> 
>      class Typeable k => Typeable (a :: k)
> 
> If I'm wrong, could someone please elaborate why?  If I'm right,
> please read on ...
> 
> That would be a breaking change, but could we at least get
> 
>      kindable :: forall (a :: k). Typeable a => Dict (Typable (a :: k))
> 
> in the meantime?
> 
> Tom

-- 
Adam Gundry, Haskell Consultant
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