minh thu wrote: > Also, I'd like to know why > > id id True > > is permitted but not > > (\f -> f f True) id
If you want to do this, answer the question "what is the type of (\f -> f f True)"? You can do this, by the way, using rank-2 types: > {-# LANGUAGE Rank2Types, PatternSignatures #-} > thisIsAlwaysTrue = (\ (f :: forall a. a -> a) -> f f True) id Cheers, Jochem -- Jochem Berndsen | joc...@functor.nl | joc...@牛在田里.com _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe