在 ST Monad 中转换的函数参数

Question

我如何编写以下函数 tt，当前存在类型错误：

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

我认为在 tt 中的 runST 中，状态变量 s 可以线程化到函数 f 中，但是下面的编译器错误告诉我我错了:

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 }'

任何意见将不胜感激。

Answer 1

伪匿名发布答案：

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

在 ST Monad 中转换的函数参数

Function argument which transforms in ST Monad

haskell

st-monad