Haskell 基础 - 模式匹配与列表尾部

Question

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

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-- Ex 5: reverse a NonEmpty list.\n--\n-- PS. The Data.List.NonEmpty type has been imported for you\n\n-- below doesn\'t work\n-- reverseNonEmpty :: NonEmpty a -> NonEmpty a\n-- reverseNonEmpty  (h:|tail) = head reversed :| (tail reversed)\n--   where \n--     reversed = reverse (h:tail)\n\n-- below works\nreverseNonEmpty :: NonEmpty a -> NonEmpty a\nreverseNonEmpty (h :| tail) = head reversed :| revTail\n  where\n    reversed = reverse (h : tail)\n    -- revTail = (tail reversed) - this doesn\'t work, but WHY???\n    (_:revTail) = reversed\n

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错误信息是：

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Set7.hs:152:15: error:\n    \xe2\x80\xa2 Couldn\'t match expected type \xe2\x80\x98[a] -> t\xe2\x80\x99 with actual type \xe2\x80\x98[a]\xe2\x80\x99\n    \xe2\x80\xa2 The function \xe2\x80\x98tail\xe2\x80\x99 is applied to one argument,\n      but its type \xe2\x80\x98[a]\xe2\x80\x99 has none\n      In the expression: tail reversed\n      In an equation for \xe2\x80\x98revTail\xe2\x80\x99: revTail = tail reversed\n    \xe2\x80\xa2 Relevant bindings include\n        revTail :: t (bound at Set7.hs:152:5)\n        reversed :: [a] (bound at Set7.hs:151:5)\n        tail :: [a] (bound at Set7.hs:149:23)\n        h :: a (bound at Set7.hs:149:18)\n        reverseNonEmpty :: NonEmpty a -> NonEmpty a\n          (bound at Set7.hs:149:1)\n    |\n152 |     revTail = tail reversed\n

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我不明白错误消息，有什么问题吗tail reversed？

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为什么具有模式匹配功能的版本(_:revTail) = reversed有效？

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Answer 1

reverseNonEmpty (h :| tail) = ...

您自己绑定了名称tail，以引用NonEmpty. 所以这会影响正常的tail函数，并且无法在 的体内引用它reverseNonEmpty。

这就是为什么错误消息告诉您它tail具有类型[a]并且不能应用于参数。