如何在Haskell中实现fromJust和遍历字符串列表
所以我得到了这些函数：
intercalate' :: [a] -> [[a]] -> [a]
intercalate' xs xss = concat (intersperse' xs xss)
intersperse' :: a -> [a] -> [a]
intersperse' _ [] = []
intersperse' sep (x:xs) = x : prependToAll' sep xs
prependToAll' :: a -> [a] -> [a]
prependToAll' _ [] = []
prependToAll' sep (x:xs) = sep : x : prependToAll' sep xs
encodeWord :: Table -> String -> Maybe Code
encodeWord table str = intercalate' [Silence, Silence]
<$> mapM (\x -> lookup x table) str对于此数据：
module Types where
data Atom = Beep | Silence
deriving (Eq, Show)
type Code = [Atom]
dit, dah, shortGap, mediumGap :: Code
dit = [Beep, Silence]
dah = [Beep, Beep, Beep, Silence]
shortGap = replicate (3-1) Silence
mediumGap = replicate (7-1) Silence
morseCode :: Char -> Code
morseCode 'A' = dit ++ dah
morseCode 'B' = dah ++ dit ++ dit ++ dit
morseCode 'C' = dah ++ dit ++ dah ++ dit
morseCode 'D' = dah ++ dit ++ dit
morseCode 'E' = dit
morseCode 'F' = dit ++ dit ++ dah ++ dit
morseCode 'G' = dah ++ dah ++ dit
morseCode 'H' = dit ++ dit ++ dit ++ dit
morseCode 'I' = dit ++ dit
morseCode 'J' = dit ++ dah ++ dah ++ dah
morseCode 'K' = dah ++ dit ++ dah
morseCode 'L' = dit ++ dah ++ dit ++ dit
morseCode 'M' = dah ++ dah
morseCode 'N' = dah ++ dit
morseCode 'O' = dah ++ dah ++ dah
morseCode 'P' = dit ++ dah ++ dah ++ dit
morseCode 'Q' = dah ++ dah ++ dit ++ dah
morseCode 'R' = dit ++ dah ++ dit
morseCode 'S' = dit ++ dit ++ dit
morseCode 'T' = dah
morseCode 'U' = dit ++ dit ++ dah
morseCode 'V' = dit ++ dit ++ dit ++ dah
morseCode 'W' = dit ++ dah ++ dah
morseCode 'X' = dah ++ dit ++ dit ++ dah
morseCode 'Y' = dah ++ dit ++ dah ++ dah
morseCode 'Z' = dah ++ dah ++ dit ++ dit
morseCode '1' = dit ++ dah ++ dah ++ dah ++ dah
morseCode '2' = dit ++ dit ++ dah ++ dah ++ dah
morseCode '3' = dit ++ dit ++ dit ++ dah ++ dah
morseCode '4' = dit ++ dit ++ dit ++ dit ++ dah
morseCode '5' = dit ++ dit ++ dit ++ dit ++ dit
morseCode '6' = dah ++ dit ++ dit ++ dit ++ dit
morseCode '7' = dah ++ dah ++ dit ++ dit ++ dit
morseCode '8' = dah ++ dah ++ dah ++ dit ++ dit
morseCode '9' = dah ++ dah ++ dah ++ dah ++ dit
morseCode '0' = dah ++ dah ++ dah ++ dah ++ dah
morseCode _ = undefined -- Avoid warnings
type Table = [(Char, Code)]
morseTable :: Table
morseTable = [ (c , morseCode c) | c <- ['A'..'Z']++['0'..'9'] ]encodeWord函数按预期工作。
示例:输入："HELLO"输出：[Beep,Silence,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Beep,Beep,Silence]
现在，我尝试定义一个新函数: encodeWords。
示例:输入：["HI","THERE"]输出：[Beep,Silence,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Silence,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Silence]
到目前为止，我已经做到了这一点。
encodeWords :: Table -> [String] -> Maybe Code
encodeWords table stringList = intercalate' [Silence, Silence, Silence,Silence,Silence,Silence]
<$> mapM (\x -> encodeWord table x ) stringList我希望最后一个函数会有类型签名：
Table -> [String] -> Code对于encodeWord，我只设法编写了Table -> [String] -> Maybe Code。我尝试过这样使用fromJust：
import Data.Maybe
encodeWord :: Table -> String -> Code
encodeWord table str = fromJust (intercalate' [Silence, Silence]
<$> mapM (\x -> lookup x table) str)但是，我只能在我正在编写的程序中使用Prelude和Data.Char。
当我尝试的时候：
fromJust :: HasCallStack => Maybe a -> a
fromJust Nothing = error "Maybe.fromJust: Nothing" -- yuck
fromJust (Just x) = x
type HasCallStack = (?callStack :: CallStack)
encodeWord :: Table -> String -> Code
encodeWord table str = fromJust (intercalate' [Silence, Silence]
<$> mapM (\x -> lookup x table) str)我刚刚得到了这个错误：
Operator applied to too few arguments: ?
|
59 | type HasCallStack = (?callStack :: CallStack)
| ^
Failed, one module loaded.有没有一种简单的方法可以让我自己实现fromJust，只需使用Data.Char和Prelude？
总之，我正在尝试自己实现fromJust，并让这个函数正常工作：
encodeWords :: Table -> [String] -> Maybe Code
encodeWords table stringList = intercalate' [Silence, Silence, Silence,Silence,Silence,Silence]
<$> mapM (\x -> encodeWord table x ) stringList我不确定是否应该使用Map、MapM或其他方法将encodeWord应用于列表中的每个字符串。之后，对于输入列表中的每个字符串，encodeWords应该在输出Code之间添加6个Silence。
回答 2
Stack Overflow用户
发布于 2020-11-17 21:12:35
如果出现错误，返回Nothing是一种完全可行的错误管理策略，您可以级联它。
最后，您总是可以将结果放入fromJust，如果有非法输入，则后果自负。正如HaskellFreak提到的，您的fromJust版本看起来还可以。
这段代码似乎可以工作：
myFromJust :: Maybe a -> a
myFromJust Nothing = error "Maybe.fromJust: Nothing" -- yuck
myFromJust (Just x) = x
encodeWord :: Table -> String -> Maybe Code
encodeWord table str = (intercalate' [Silence, Silence]
<$> mapM (\x -> lookup x table) str)在ghci下进行测试
λ>
λ> encodeWord morseTable "ABC"
Just [Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Silence]
λ>
λ>
λ> encodeWord morseTable "ABC+"
Nothing
λ>现在尝试编写更高级别的函数encodeWords
如果我们从一个简单的map构造开始：
λ>
λ> stringList = ["ATTACK","AT","DAWN"]
λ>
λ> :type (map (encodeWord morseTable) stringList)
(map (encodeWord morseTable) stringList) :: [Maybe Code]
λ>所以我们有一个可能的Code对象。考虑到intercalate'的类型签名，我们倾向于使用Maybe [Code]签名。
λ>
λ> :type mapM
mapM :: (Traversable t, Monad m) => (a -> m b) -> t a -> m (t b)
λ> 因此，如果类型a与m b相同，并且最左边的函数参数为id，我们可以看到mapM可以用可能的list结构交换list结构。例如：(m b -> m b) -> t (m b) -> m (t b)。
λ>
λ> :type (mapM id)
(mapM id) :: (Traversable t, Monad m) => t (m b) -> m (t b)
λ>
λ> :type (mapM id $ map (encodeWord morseTable) stringList)
(mapM id $ map (encodeWord morseTable) stringList) :: Maybe [Code]
λ>
λ>
λ> sil6 = replicate 6 Silence
λ>
λ> :type ((intercalate' sil6) <$> mapM id (map (encodeWord morseTable) stringList))
((intercalate' sil6) <$> mapM id (map (encodeWord morseTable) stringList))
:: Maybe [Atom]
λ> 因此，这可能是encodeWords的有效代码
encodeWords :: Table -> [String] -> Maybe Code
encodeWords table stringList =
let sil6 = replicate 6 Silence
in
intercalate' sil6
<$> (mapM id $ map (encodeWord table) stringList)在ghci下进行测试
λ>
λ> length $ myFromJust (encodeWords morseTable stringList)
112
λ> (encodeWords morseTable stringList)
Just [Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Silence,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Silence,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Silence,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Silence,Beep,Beep,Beep,Silence,Beep,Beep,Beep,Silence,Silence,Silence,Beep,Beep,Beep,Silence,Beep,Silence]
λ> 请注意，在该上下文中，mapM id与库函数sequence :: Monad m => t (m a) -> m (t a)是相同的，后者肯定在前缀集中。
为了完全公开，Prelude库还包括了一个traverse函数，这是sequence和map之间的一种调情。因此，您可以像这样编写一个稍微更优雅的encodeWords版本：
encodeWords2 :: Table -> [String] -> Maybe Code
encodeWords2 table stringList =
let sil6 = replicate 6 Silence
in intercalate' sil6 <$> traverse (encodeWord table) stringListStack Overflow用户
发布于 2020-11-17 20:02:56
你写的fromJust的定义直接来自于Data.Maybe。如果不进行简单的修改，这将无法工作：
fromJust :: Maybe a -> a
fromJust Nothing = error "fromJust: Nothing"
fromJust (Just x) = x正如您所看到的，fromJust模式匹配Maybe数据类型的构造函数，就像您对morseCode所做的那样。
有关模式匹配的详情，请参阅：http://learnyouahaskell.com/syntax-in-functions
https://stackoverflow.com/questions/64881810
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