## Why are higher rank types so fragile in Haskell

问题 I was messing around with the runST function. Which has type (forall s. ST s a) -> a and it seems like trying to use it in any way that isn't directly applying without any indirection breaks it in pretty nasty ways. runST :: (forall s. ST s a) -> a const :: a -> b -> a so by substituting a in const for forall s. ST s a you should get the type of const runST const runST :: b -> (forall s. ST s a) -> a but instead GHC says that it can't match a with (forall s. ST s a) -> a but since a literally