为什么 GHCi 没有在此错误消息中显示多态类型？

Question

我将类型为 Num a => a -> a 的变量的类型归因于 Int，这会按预期引发错误。但是，错误并不完全符合我的预期。

GHCi

λ> let x = typeInference 1
λ> :t x
x :: Num a => a -> a
λ> x :: Int

<interactive>:141:1: error:
    • Couldn't match expected type 'Int'
                  with actual type 'Integer -> Integer'
    • Probable cause: 'x' is applied to too few arguments
      In the expression: x :: Int
      In an equation for 'it': it = x :: Int
λ>

类型推理定义

typeInference :: Num a => a -> a -> a
typeInference x y = x + y + 1

我以为错误消息会说 with actual type 'Num a => a -> a'，这是多态类型，为什么不是呢？这与 GHCi 的默认类型有关吗？

Answer 1

这确实是因为GHCi's type defaulting.

相关规则在 section 4.3.4 of the Haskell 2010 Report 中描述，这部分特别相关：

Each defaultable variable is replaced by the first type in the default list that is an instance of all the ambiguous variable’s classes. It is a static error if no such type is found.

每个可默认的变量都被替换为它的默认值，这是在产生错误消息之前发生的操作。

当类型不明确时，类型默认必须发生在类型检查之前。特别是，如果保留类型 class 约束而不是默认约束，则表达式 x :: Int 中的类型将有歧义，如

中所定义

We say that an expression e has an ambiguous type if, in its type ∀ u'. cx ⇒ t, there is a type variable u in u' that occurs in cx but not in t. Such types are invalid.

（摘自 Haskell 2010 年报告，将 u 替换为带 u' 的上划线）。

因为，如果类型有效，则整个表达式将有一个类型变量（受 Num 约束的变量）不会出现在结果类型 (Int) 中（并且type variable cannot be unified away), defaulting must be occur here before the type checking.

这可以通过意识到我们正在进行类型检查而变得更明确一些

(x :: forall a. Num a => a -> a) :: Int

看起来，在统一期间，如果最外层的类型构造函数不匹配（(->)在左侧，Int在右侧），它必须默认，因为它不能自动深入统一（如果右侧的最外层类型构造函数也是 (->)）。

以下是我测试过的几个遵循此行为的示例：

ghci> :set -XExplicitForAll
ghci> (x :: forall a. Num a => a -> a) :: Char -> Char  -- Outermost constructor matches, so 'a' can get unified with Char and the 'a' type variable disappears

<interactive>:5:2: error:
    • No instance for (Num Char)
        arising from an expression type signature
    • In the expression:
          (x :: forall a. Num a => a -> a) :: Char -> Char
      In an equation for ‘it’:
          it = (x :: forall a. Num a => a -> a) :: Char -> Char
ghci> (x :: forall a. Num a => a -> a) :: Maybe Char

<interactive>:8:2: error:
    • Couldn't match expected type ‘Maybe Char’
                  with actual type ‘Integer -> Integer’
    • In the expression: (x :: forall a. Num a => a -> a) :: Maybe Char
      In an equation for ‘it’:
          it = (x :: forall a. Num a => a -> a) :: Maybe Char
ghci> (x :: forall a. Num a => a -> a) :: Either Char Bool

<interactive>:10:2: error:
    • Couldn't match expected type ‘Either Char Bool’
                  with actual type ‘Integer -> Integer’
    • In the expression:
          (x :: forall a. Num a => a -> a) :: Either Char Bool
      In an equation for ‘it’:
          it = (x :: forall a. Num a => a -> a) :: Either Char Bool
ghci> (x :: forall a. Num a => (,) a a) :: Either Char Bool

<interactive>:11:2: error:
    • Couldn't match expected type ‘Either Char Bool’
                  with actual type ‘(Integer, Integer)’
    • In the expression:
          (x :: forall a. Num a => (,) a a) :: Either Char Bool
      In an equation for ‘it’:
          it = (x :: forall a. Num a => (,) a a) :: Either Char Bool
ghci> (x :: forall a. Num a => (,) a a) :: Char

<interactive>:12:2: error:
    • Couldn't match expected type ‘Char’
                  with actual type ‘(Integer, Integer)’
    • In the expression: (x :: forall a. Num a => (,) a a) :: Char
      In an equation for ‘it’:
          it = (x :: forall a. Num a => (,) a a) :: Char

为什么 GHCi 没有在此错误消息中显示多态类型？

How come GHCi doesn't display the polymorphic type in this error message?

haskell

types

ghci