함수 인수

Question

I 현재 입력 한 다음 함수 tt를 작성할 수있는 방법 오류 :

I가 tt에 runST, 내부 s 가변 상태 함수 f에 나사 결합 될 수 있다는 것을 생각

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 

, 하지만 컴파일러 오류가 다음과 같이 나에게 잘못되었다고 알려줍니다.

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 }' 

모든 의견은 깊이 감사하겠습니다.

Answer 1

대답은 의사 익명으로 게시했습니다 :

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