[Haskell-cafe] forall in a lambda abstraction

[Scott J.]

Hello,

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

Thanks
Jan


id x = x
test1 = \f ::(forall a. a -> a) -> (f True,f 'a')  
test = \f -> (f True,f False)
h :: (forall a. a->a) -> (Bool,Char)
--h :: (a.a -> a) -> (Bool,Char)
h id = (id True, id 'a')
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