haskell - 如何报告 Haskell 代码中的错误

Question

我正在编写一个程序，该程序返回列表中的每个第 r 个元素。该列表可以是任何类型。我想在 r 为零时报告错误，但我的代码无法正常工作(当我注释掉错误行时，它工作正常)。谁能告诉我在这种情况下如何报告错误

rthElem :: Int -> [a] -> [a]
rthElem _ [] = []
rthElem 0 (x:xs) = "Error"
rthElem n (x:xs) = rthElem' n 1 (x:xs) where
    rthElem' n i (x:xs)  = (if (n `divides` i) then
        [x] else
        [])
        ++ (rthElem' n (i+1) xs)
    rthElem' _ _ [] = []
    divides x y = y `mod` x == 0  

Answer 1

在这种情况下，您可以使用Maybe或Either。

这就是也许的样子。 没有什么会成为我们的“错误”。

rthElem :: Int -> [a] -> Maybe [a]
rthElem _ [] = Just []
rthElem 0 (x:xs) = Nothing
rthElem n (x:xs) = Just (rthElem' n 1 (x:xs)) where
    rthElem' n i (x:xs)  = (if (n `divides` i) then
        [x] else
        [])
        ++ (rthElem' n (i+1) xs)
    rthElem' _ _ [] = []
    divides x y = y `mod` x == 0

main :: IO ()
main = case (rthElem 0 [1..5]) of
  Nothing  -> putStrLn "Error"
  Just elm -> print elm

另一种方法是使用Either。 Either 将返回 Left 或 Right。 左边将是我们的“错误”。

rthElem :: Int -> [a] -> Either String [a]
rthElem _ [] = Right []
rthElem 0 (x:xs) = Left "Error"
rthElem n (x:xs) = Right (rthElem' n 1 (x:xs)) where
    rthElem' n i (x:xs)  = (if (n `divides` i) then
        [x] else
        [])
        ++ (rthElem' n (i+1) xs)
    rthElem' _ _ [] = []
    divides x y = y `mod` x == 0

main :: IO ()
main = case (rthElem 0 [1..5]) of
  Left err  -> putStrLn err
  Right elm -> print elm

最好的方法是使用Either。了解有关错误处理的更多信息 here .