为什么Haskell不能推导出这种类型

Question

我一直在尝试编写一个程序来实现任意域上的多项式，一种数学结构。我选择了Haskell作为编程语言，我使用了GADTS语言扩展。但是，我不明白为什么GHCi不能推导出A的约束条件。

-- 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)

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)
   |            

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是字段的实例，这意味着A是环的实例。所以调用fzero::a就像调用42::int一样，应该是合理的。此外，我已经编写了showa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的实例。

显然，译员的想法不同。我哪里搞错了？

Answer 1

从Arrowd的评论来看：

代码很好，但需要ScopedTypeVariables扩展名，这使得FZero::A中的类型变量A引用前面介绍的A。