Re: [Haskell-cafe] type inference question

8 Oct 2009


      2009/10/8 Jochem Berndsen :
...
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
So I learned we should be explicit about f being used polymorphically;
that's what rank-2 types are for. (correct ?)

Thanks,
Thu