Haskell 为什么可以’；哈斯克尔推导出这种类型

我一直在试图写一个程序来实现任意领域的多项式，一个数学结构。我选择Haskell作为编程语言，并使用了

GADTs

语言扩展。然而，我不明白为什么GHCi不能推断出

a

的约束

背景：

-- irreducible.hs

{-# LANGUAGE GADTs #-}

infixl 6 .+
infixl 7 .*

class Ring a where
  (.+) :: a -> a -> a
  (.*) :: a -> a -> a
  fneg :: a -> a
  fzero :: a
  funit :: a

class (Ring a) => Field a where
  finv :: a -> a

data Polynomial a where 
  Polynomial :: (Field a) => [a] -> Char -> Polynomial a

instance (Show a) => Show (Polynomial a) where
  show (Polynomial (a0:ar) x)
    = show a0
      ++ concatMap (\(a, k) -> "+" ++ show a ++ x:'^':show k) (zip ar [0..])
  show (Polynomial [] _) = show (fzero::a)

说明：环是定义了加法和乘法的东西，加法形成一个（实际上是阿贝尔的）群，乘法形成一个幺半群。域是定义了乘法逆的环。域上的多项式由系数列表和字符表示。字符，例如

'x'

，表示该多项式与未知变量

x

有关。对于写为

多项式[]'x'

的零多项式，我希望它显示底层字段的零元素

在运行GHCi后，我得到以下信息：

irreducible.hs:59:28: error:
    • Could not deduce (Show a0) arising from a use of ‘show’
      from the context: Show a
        bound by the instance declaration at irreducible.hs:55:10-40
      or from: Field a
        bound by a pattern with constructor:
                   Polynomial :: forall a. Field a => [a] -> Char -> Polynomial a,
                 in an equation for ‘show’
        at irreducible.hs:59:9-23
      The type variable ‘a0’ is ambiguous
      These potential instances exist:
        instance (Show a, Show b) => Show (Either a b)
          -- Defined in ‘Data.Either’
        instance Show Ordering -- Defined in ‘GHC.Show’
        instance Show Integer -- Defined in ‘GHC.Show’
        ...plus 25 others
        ...plus 87 instances involving out-of-scope types
        (use -fprint-potential-instances to see them all)
    • In the expression: show (fzero :: a)
      In an equation for ‘show’:
          show (Polynomial [] _) = show (fzero :: a)
      In the instance declaration for ‘Show (Polynomial a)’
   |
59 |   show (Polynomial [] _) = show (fzero::a)
   |                            ^^^^^^^^^^^^^^^

irreducible.hs:59:34: error:
    • Could not deduce (Ring a1) arising from a use of ‘fzero’
      from the context: Show a
        bound by the instance declaration at irreducible.hs:55:10-40
      or from: Field a
        bound by a pattern with constructor:
                   Polynomial :: forall a. Field a => [a] -> Char -> Polynomial a,
                 in an equation for ‘show’
        at irreducible.hs:59:9-23
      Possible fix:
        add (Ring a1) to the context of
          an expression type signature:
            forall a1. a1
    • In the first argument of ‘show’, namely ‘(fzero :: a)’
      In the expression: show (fzero :: a)
      In an equation for ‘show’:
          show (Polynomial [] _) = show (fzero :: a)
   |
59 |   show (Polynomial [] _) = show (fzero::a)
   |            

现在，让我们关注有问题的部分：

instance (Show a) => Show (Polynomial a) where
  show (Polynomial (a0:ar) x) = show a0 ++ [...]
  show (Polynomial [] _) = show (fzero::a)

在我看来，

多项式a

保证

a

是

字段

的一个实例，这意味着

a

是

环

的一个实例。因此，调用

fzero:：a

，就像调用

42:：Int

，应该是合理的。此外，我已经编写了

Show a

作为约束，并且

多项式a

的构造函数具有

多项式[a]Char

的形状，因此它还应该知道

a0

的类型是

Show

的一个实例

显然，译员的想法不同。我在哪里犯了错误？

来自arrowd的评论：

---

代码很好，但需要扩展名

ScopedTypeVariables

，这使得

fzero:：a中的类型变量
a
参考前面介绍的
a
常见错误，搜索
ScopedTypeVariables
@arrowd谢谢！这对我很有用。