[Haskell-cafe] RankNTypes question

[Nikolay Amiantov]

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.