On Saturday 21 May 2011 14:55:54, Paolo G. Giarrusso wrote: > However, it turns out that > Prelude> let id2 :: Int -> Int = \x -> x > works but there's no way whatsoever to make the following work: > Prelude> let (id2 :: forall t. t -> t) = \x -> x
Works without problems or extensions in the form Prelude> let id2 :: t -> t; id2 = \x -> x Prelude> :t id2 id2 :: t -> t Also Prelude> let a :: t -> t; a = a a Prelude> :i a a :: t -> t -- Defined at <interactive>:1:18 (of course, using a loops). _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
