Correction: I meant
data Baz f a = Baz (Foo f => f a)
(Dropped the 'forall', which would make the inner 'f' have nothing to do with
the type parameter 'f' of 'Baz')
On 2011-June-09 Thursday 01:07:13 Daniel Schüssler wrote:
> Hello,
>
> you might be thinking of this type?
>
> {-# LANGUAGE Rank2Types #-}
>
> class Foo f where
> foo :: a -> f a
>
> data Baz f a = Baz (forall f. Foo f => f a)
>
> instance Foo (Baz f) where
> foo a = Baz (foo a)
>
> Maybe the difference between Bar and Baz ist best explained by writing it
> with an explicit class dictionary for Foo:
>
> {-# LANGUAGE Rank2Types #-}
>
> data FooDict f = FooDict {
> foo :: forall a. a -> f a
> }
>
> data Bar f a = Bar (FooDict f) (f a)
>
> data Baz f a = Baz (FooDict f -> f a)
>
> fooDict_Baz :: FooDict (Baz f)
> fooDict_Baz = FooDict (\a -> Baz (\d -> foo d a))
>
> -- fooDict_Bar :: FooDict (Bar f)
> -- fooDict_Bar = FooDict (\a -> Bar ? ?)
> -- Doesn't work - you'd have to create a 'FooDict f' and a 'f a' out of
> just an 'a'
>
>
>
> Cheers,
> Daniel
>
> On 2011-June-08 Wednesday 20:45:56 Guy wrote:
> > {- continuing discussion from beginners@ -}
> >
> > I have code such as
> >
> > class Foo f where
> >
> > foo :: a -> f a
> >
> > data Bar f a = Foo f => Bar {bar :: f a}
> >
> > instance Foo (Bar f) where
> >
> > foo a = Bar $ foo a
> >
> > GHC insists that I put Foo f => on the instance declaration, even though
> > the constructor for Bar implies this.
> >
> > Is there any reason why GHC cannot infer this constraint from the Bar
> > constructor? One issue raised in the beginners thread is that
> > undefined :: Bar f a
> > is not Foo f, but as undefined cannot be evaluated, this would not appear
> > to be a problem.
> >
> >
> > _______________________________________________
> > Haskell-Cafe mailing list
> > Haskell-Cafe@haskell.org
> > http://www.haskell.org/mailman/listinfo/haskell-cafe
>
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