오류 t ~ t -> T1

Question

나는이 다음 코드 : 나는 GHCi에서 다음과 같은 오류가 점점 오전

{-# LANGUAGE DeriveFunctor #-} 

data Foo a = Foo { field1 :: a, field2 :: a} deriving (Functor) 

instance Applicative Foo where 
    pure a = Foo a a 
    f <*> a = Foo (r field1) (r field2) 
    where r g = g f $ g a 

:

help.hs:8:23: error: 
    • Occurs check: cannot construct the infinite type: t ~ t -> t1 
    • In the second argument of ‘($)’, namely ‘g a’ 
     In the expression: g f $ g a 
     In an equation for ‘r’: r g = g f $ g a 
    • Relevant bindings include 
     g :: Foo (a -> b) -> t -> t1 (bound at help.hs:8:13) 
     r :: (Foo (a -> b) -> t -> t1) -> t1 (bound at help.hs:8:11) 

help.hs:8:25: error: 
    • Couldn't match type ‘a’ with ‘a -> b’ 
     ‘a’ is a rigid type variable bound by 
     the type signature for: 
      (<*>) :: forall a b. Foo (a -> b) -> Foo a -> Foo b 
     at help.hs:7:5 
     Expected type: Foo (a -> b) 
     Actual type: Foo a 
    • In the first argument of ‘g’, namely ‘a’ 
     In the second argument of ‘($)’, namely ‘g a’ 
     In the expression: g f $ g a 
    • Relevant bindings include 
     g :: Foo (a -> b) -> t -> t1 (bound at help.hs:8:13) 
     r :: (Foo (a -> b) -> t -> t1) -> t1 (bound at help.hs:8:11) 
     a :: Foo a (bound at help.hs:7:9) 
     f :: Foo (a -> b) (bound at help.hs:7:3) 
     (<*>) :: Foo (a -> b) -> Foo a -> Foo b (bound at help.hs:7:3) 

어떻게 Foo의이 적용 가능한 Functor 인스턴스를 구현하여 컴파일 할 수 있습니까?

Answer 1

r g = g f $ g a 

다형성 함수로 g 필요한 정의 - GHC 이러한 복합 타입을 추론 할 수 없다.

당신은 당신이 정말로 당신의 코드를 컴파일하려면, 당신의 유형에 더 명시해야합니다

instance Applicative Foo where 
    pure a = Foo a a 
    (Foo f1 f2) <*> (Foo a1 a2) = Foo (f1 a1) (f2 a2)