PR Stanley <[EMAIL PROTECTED]> writes: > It's one of those things - I know sort of instinctively why it is so > but can't think of the formal rationale for it:
> f g x = g (g x) :: (t -> t) -> (t -> t) (t -> t) -> (t -> t) So g :: t -> t x :: t Thus f :: (t -> t) -> t -> t (The last parenthesis is not necessary, but implies that the type of the partial application f g is a function t -> t .) -k -- If I haven't seen further, it is by standing in the footprints of giants _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
