[Haskell-beginners] [Haskell-cafe] arbitrary rank polymorphism and ghc language pragmas

[rickmurphy]

Thanks Francesco. And I did verify that ExplicitForAll does in fact
allow Rank 1 Types in functions like the following ...

f :: (forall a. a -> a)

--
Rick

On Thu, 2012-07-05 at 16:28 +0100, Francesco Mazzoli wrote:
> At Thu, 05 Jul 2012 11:18:00 -0400,
> rickmurphy  wrote:
> > data T = TC (forall a b. a -> b -> a)
> 
> The type of `TC' will be `(forall a b. a -> b -> a) -> T', a Rank-2
> type.
> 
> --
> Francesco * Often in error, never in doubt
> 
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