Haskell 基础 - 模式匹配与列表尾部

Question

我正在做 Set7.hs 来自 haskell.mooc.fi/ Exercises7

-- Ex 5: reverse a NonEmpty list.
--
-- PS. The Data.List.NonEmpty type has been imported for you

-- below doesn't work
-- reverseNonEmpty :: NonEmpty a -> NonEmpty a
-- reverseNonEmpty  (h:|tail) = head reversed :| (tail reversed)
--   where 
--     reversed = reverse (h:tail)

-- below works
reverseNonEmpty :: NonEmpty a -> NonEmpty a
reverseNonEmpty (h :| tail) = head reversed :| revTail
  where
    reversed = reverse (h : tail)
    -- revTail = (tail reversed) - this doesn't work, but WHY???
    (_:revTail) = reversed

错误信息是：

Set7.hs:152:15: error:
    • Couldn't match expected type ‘[a] -> t’ with actual type ‘[a]’
    • The function ‘tail’ is applied to one argument,
      but its type ‘[a]’ has none
      In the expression: tail reversed
      In an equation for ‘revTail’: revTail = tail reversed
    • Relevant bindings include
        revTail :: t (bound at Set7.hs:152:5)
        reversed :: [a] (bound at Set7.hs:151:5)
        tail :: [a] (bound at Set7.hs:149:23)
        h :: a (bound at Set7.hs:149:18)
        reverseNonEmpty :: NonEmpty a -> NonEmpty a
          (bound at Set7.hs:149:1)
    |
152 |     revTail = tail reversed

我不明白错误消息，这是怎么回事

tail reversed

？

为什么带有模式匹配的版本

(_:revTail) = reversed

有效？

Answer 1

reverseNonEmpty (h :| tail) = ...

您自己绑定了名称

tail

，以引用

NonEmpty

的列表部分。因此，这遮盖了正常的

tail

函数，并且无法在

reverseNonEmpty

的体内引用该函数。

这就是为什么错误消息告诉您

tail

的类型为

[a]

并且不能应用于参数。