我想拿一个字符串然后变成一个列表Direction.例如,"UDDUD"应该返回[U,D,D,U,D],而任何不包含U或D返回的字符串Nothing(例如"UDYD"返回Nothing).
data Direction = U | D
deriving (Show, Eq)
-- where U is Up and D is Down
findDirection :: [Char] -> Maybe [Direction]
findDirection [] = Nothing
findDirection ['U'] = Just [U]
findDirection ['D'] = Just [D]
findDirection (x:xs)
| x == 'U' = Just (U : findDirection xs)
| x == 'D' = Just (D : findDirection xs)
| otherwise = Nothing
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我收到以下错误:
Couldn't match expected type ‘[Direction]’
with actual type ‘Maybe [Direction]’
In the second argument of ‘(:)’, namely ‘findDirection xs’
In the first argument of ‘Just’, namely
‘(U : findDirection xs)’
Test_findDirection.hs:8:32:
Couldn't match expected type ‘[Direction]’
with actual type ‘Maybe [Direction]’
In the second argument of ‘(:)’, namely ‘findDirection xs’
In the first argument of ‘Just’, namely
‘(D : findDirection xs)’
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据我了解,Just (D : findDirection xs)和Just (U : findDirection xs)有型的[Direction]?为什么会这样?我做错了什么,在这里?
Just (D : findDirection xs)并且Just (U : findDirection xs)是[方向]类型?为什么会这样?我做错了什么,在这里?
不,Just (D : findDirection xs)实际上是不合格的.让我们剖析这个错误信息:
Couldn't match expected type ‘[Direction]’
with actual type ‘Maybe [Direction]’
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我们正在使用Maybe [Direction]我们应该使用的地方[Direction].
In the second argument of ‘(:)’, namely ‘findDirection xs’
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啊哈.我们用(:) :: a -> [a] -> [a]错了.毕竟,findDirection将返回一个Maybe [Direction],而不是一个[Direction].我们需要这样的东西:
consOnMaybe :: a -> Maybe [a] -> Maybe [a]
consOnMaybe _ Nothing = Nothing
consOnMaybe x (Just xs) = Just (x : xs)
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现在你的功能可以写成
findDirection (x:xs)
| x == 'U' = consOnMaybe U (findDirection xs)
| x == 'D' = consOnMaybe D (findDirection xs)
| otherwise = Nothing
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或者,我们可以使用consOnMaybe x = fmap (x:).作为额外的奖励,这里是一个使用预定义函数而没有显式递归的变体(练习:了解它是如何工作的)
findDirection :: [Char] -> Maybe [Direction]
findDirection [] = Nothing
findDirection xs = traverse toDirection xs
toDirection :: Char -> Maybe Direction
toDirection 'U' = Just U
toDirection 'D' = Just D
toDirection _ = Nothing
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