Haskell：类型变量和泛型类型

Question

当我在函数的类型签名中明确提到 isQuestion 类型时，GHCi 完美地编译了它：

isQuestion :: [Char] -> Maybe Bool
isQuestion [] = Nothing
isQuestion xs = Just (last xs == '?')

但是，当我转向 'generic' 代码时，它不起作用：

isQuestion :: [a] -> Maybe b
isQuestion [] = Nothing
isQuestion xs = Just (last xs == '?')

因为我得到以下错误：

<interactive>:138:17: error:
    * Couldn't match type `b' with `Bool'
      `b' is a rigid type variable bound by
        the type signature for:
          isQuestion :: forall a b. [a] -> Maybe b
        at <interactive>:136:1-28
      Expected type: Maybe b
        Actual type: Maybe Bool
    * In the expression: Just (last xs == '?')
      In an equation for `isQuestion':
          isQuestion xs = Just (last xs == '?')
    * Relevant bindings include
        isQuestion :: [a] -> Maybe b (bound at <interactive>:137:1)

Answer 1

对于 forall a b. [a] -> Maybe b，类型 [a] -> Maybe b 是 shorthand。但是 isQuestion 不适用于所有类型 a 和 b，它仅在 a 是 [=16= 时才有效] 并且 b 是 Bool.

Answer 2

参数多态性不是子类型或超类型。它声明值的实现不依赖于部分类型。事实上，该实现适用于任何类型的选择。

您的实施不适用于任何类型选择。它仅在 a 为 Char 且 b 为 Bool 时有效。你可以看出这一点，因为它在那里使用了 (== '?')，这是一个 Char -> Bool.

类型的函数

Answer 3

第一个观察结果是

last xs == something

只有在 xs 的元素范围内有 (==) 的定义时才有效。但是由于编译器对 a 一无所知，所以没有这样的东西。您必须将 a 缩小为具有相等性的类型的子集：

isQuestion :: Eq a => [a] -> Maybe b

第二个观察结果是 something 在您的代码 ('?') 中是一个 Char，因此此方法只能在 a ≡ Char 时起作用。相反，您可以将其添加为参数：

isQuestion :: Eq a => a -> [a] -> Maybe b

最后，如前所述，您有一个具体的 return 类型，即 Maybe Bool，如

(==) :: a -> a -> Bool

所以你的函数的签名可以是

isQuestion :: Eq a => a -> [a] -> Maybe Bool

为清楚起见编辑了这一段请注意，根据您的业务逻辑，您可能不想对空字符串进行特殊处理。如果您唯一关心的是字符串是否以问号结尾，那么 Nothing 和 Just false 实际上是同一个意思。在这种情况下，你的函数变成了一个 "yes or no" 问题，你可以删除 Maybe:

isQuestion :: Eq a => a -> [a] -> Bool
isQuestion x = isSuffixOf [x]

或者干脆

isQuestion :: String -> Bool
isQuestion = isSuffixOf "?"

Haskell：类型变量和泛型类型

Haskell: type variables and generic types

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