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 > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
