haskell - 与中缀一起使用时，约束错误中的非类型变量参数

Question

不幸的是，我遇到了一个奇怪的错误。在将中缀与数据构造函数一起使用时会发生这种情况。

我是 Haskell 新手。任何人都可以在这方面帮助我吗？

Prelude> data L a = Cons a (L a) | Emp deriving Show
Prelude> 10 `Cons` Emp
Cons 10 Emp
Prelude> 10 `Cons` 10 `Cons` Emp

<interactive>:43:1: error:
    • Non type-variable argument in the constraint: Num (L a)
      (Use FlexibleContexts to permit this)
    • When checking the inferred type
        it :: forall a. (Num a, Num (L a)) => L (L a)
Prelude> 10 `Cons` (10 `Cons` Emp) 
Cons 10 (Cons 10 Emp)
Prelude> 10 `Cons` 10 `Cons` Emp

<interactive>:45:1: error:
    • Non type-variable argument in the constraint: Num (L a)
      (Use FlexibleContexts to permit this)
    • When checking the inferred type
        it :: forall a. (Num a, Num (L a)) => L (L a)
Prelude> data L a = Emp | Cons a (L a)  deriving Show
Prelude> 10 `Cons` 10 `Cons` Emp

<interactive>:47:1: error:
    • Non type-variable argument in the constraint: Num (L a)
      (Use FlexibleContexts to permit this)
    • When checking the inferred type
        it :: forall a. (Num a, Num (L a)) => L (L a)
Prelude> 

Answer 1

您可以将数据构造函数定义为右关联，

infixr 5 `Cons`

data L a = Cons a (L a) | Emp deriving Show

然后它将按您的预期工作。

main = print $ (21::Int) `Cons` 42 `Cons` Emp

-- => Cons 21 (Cons 42 Emp)