Haskell：功能上的非详尽模式

2 回答

• 2

  你没有处理这个案子：

  longer [] _ = undefined
  

  模式 (_:[]) 假定列表中至少有一个元素 . 在您的代码中，当第一个列表为空且第二个列表可能不为空时，您将错过这种情况 .

  回复于 2024-04-26T15:42:29+08:00
• 4

  您需要4个案例，但您不需要将两个单例列表视为单独的案例 .

  longer :: [a] -> [a] -> Bool
  
  -- 1) Two empty lists
  longer [] [] = False
  -- 2) An non-empty list and an empty list
  longer _ []  = True
  -- 3) An empty list and a non-empty list
  longer [] _ = ???
  -- 4) Two non-empty lists
  longer (_:xs) (_:ys) = longer xs ys
  

  实际上，根据 longer [] _ 应该是什么，你只需要正确的3个案例 .

  -- First case: if longer [] _ is suppose to be True
  longer :: [a] -> [a] -> Bool
  longer [] [] = True
  longer (_:xs) (_:ys) = longer xs ys
  -- We get this far if one is empty and the other is not,
  -- but we don't care which one is which.
  longer _ _ = False
  
  -- Second case: if longer [] _ is supposed to be False
  longer :: [a] -> [a] -> Bool
  longer (_:xs) (_:ys) = longer xs ys
  longer _ [] = True
  longer [] _ = False -- This covers longer [] [] as well.
  

  回复于 2024-04-26T15:42:29+08:00