Re: [Haskell-cafe] type inference question

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 | jochem@functor.nl | jochem@牛在田里.com