Ah, thanks! For some reason I haven't even thought of it -- I knew that `id` was wrong because of the type, but had no idea what else to try. Feeling stupid now. On 02/26/2015 09:46 PM, Marcin Mrotek wrote:

Well still, id is (a -> a). So given some function (forall b. b -> b) it's result is a function (forall b. b -> b). This works:

test :: (forall a. a -> a) -> Int -> Int test f a = f a

Kind regards, Marcin Mrotek

-- Nikolay.