Re: [Haskell-cafe] forall in a lambda abstraction

"Scott J." writes:

what's wrong with the code of test1 below. Ghci does complain with an error message that I don't understand.

test1 = \f ::(forall a. a -> a) -> (f True,f 'a')

test1 = \(f :: forall a. a -> a) -> (f True,f 'a')