Аргумент функции, который преобразуется в ST Monad

Question

Как я могу написать следующую функцию tt, который в настоящее время имеет тип ошибки:

t :: Int
t = runST $ do
  ref <- newSTRef 10
  readSTRef ref

tt :: (STRef s a -> ST s a) -> Int
tt f = runST $ do
  ref <- newSTRef 10
  f ref

ttTest = tt readSTRef

Я думал что внутри runST в ttпеременная состояния s можно вставить в функцию f, но следующая ошибка компилятора говорит мне, что я не прав:

transform.hs:50:3: Couldn't match type `s' with `s1' …
  `s' is a rigid type variable bound by
      the type signature for tt :: (STRef s a -> ST s a) -> Int
      at transform.hs:47:7
  `s1' is a rigid type variable bound by
       a type expected by the context: ST s1 Int
       at transform.hs:48:8
Expected type: ST s1 Int
  Actual type: ST s a
Relevant bindings include
  ref :: STRef s1 a
    (bound at transform.hs:49:3)
  f :: STRef s a -> ST s a
    (bound at transform.hs:48:4)
  tt :: (STRef s a -> ST s a) -> Int
    (bound at transform.hs:48:1)
In a stmt of a 'do' block: f ref
In the second argument of `($)', namely
  `do { ref <- newSTRef 10;
        f ref }'
transform.hs:50:3: Couldn't match type `a' with `Int' …
  `a' is a rigid type variable bound by
      the type signature for tt :: (STRef s a -> ST s a) -> Int
      at transform.hs:47:7
Expected type: ST s1 Int
  Actual type: ST s a
Relevant bindings include
  ref :: STRef s1 a
    (bound at transform.hs:49:3)
  f :: STRef s a -> ST s a
    (bound at transform.hs:48:4)
  tt :: (STRef s a -> ST s a) -> Int
    (bound at transform.hs:48:1)
In a stmt of a 'do' block: f ref
In the second argument of `($)', namely
  `do { ref <- newSTRef 10;
        f ref }'

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Answer 1

Ответ был выложен псевдоанонимно:

tt :: (forall s . STRef s Int -> ST s Int) -> Int,