Haskell 如何将列表转换为可能列表
如果我输入“L1TDR2”，我想把它改成一个[L1，Top，Down，R2]列表。但它看起来失败了Haskell 如何将列表转换为可能列表,haskell,monads,Haskell,Monads,如果我输入“L1TDR2”，我想把它改成一个[L1，Top，Down，R2]列表。但它看起来失败了 readtree :: Char -> Char -> Maybe Tree readtree x y = case x of 'L' -> case y of '1' -> Just L1 '2' -> Just L2 _ -> Nothing 'R' -> case y of
readtree :: Char -> Char -> Maybe Tree
readtree x y = case x of
'L' -> case y of
'1' -> Just L1
'2' -> Just L2
_ -> Nothing
'R' -> case y of
'1' -> Just R1
'2' -> Just R2
_ -> Nothing
'T' -> Just Top
'D' -> Just Down
_ -> Nothing
treelist :: String -> Maybe [Tree]
treelist = mapM readtree
Couldn't match type ‘Char -> Maybe Tree’ with ‘Maybe Tree’
Expected type: Char -> Maybe Tree
Actual type: Char -> Char -> Maybe Tree
Probable cause: ‘readtree’ is applied to too few arguments
In the first argument of ‘mapM’, namely ‘readtree’
In the expression: mapM readtree
x $: xs = (x:) <$> xs
readTrees ('L':'1':rest) = L1 $: readTrees rest
readTrees ('T':rest) = Top $: readTrees rest
readTrees [] = Just []
readTrees _ = Nothing
x$：xs=（x:）xs
readTrees（'L'：'1'：rest）=L1$：readTrees rest
readTrees（'T'：rest）=Top$：readTrees rest
readTrees[]=Just[]
readTrees=无
Couldn't match type ‘Char -> Maybe Tree’ with ‘Maybe Tree’
Expected type: Char -> Maybe Tree
Actual type: Char -> Char -> Maybe Tree
Probable cause: ‘readtree’ is applied to too few arguments
In the first argument of ‘mapM’, namely ‘readtree’
In the expression: mapM readtree
x $: xs = (x:) <$> xs
readTrees ('L':'1':rest) = L1 $: readTrees rest
readTrees ('T':rest) = Top $: readTrees rest
readTrees [] = Just []
readTrees _ = Nothing
x$：xs=（x:）xs
readTrees（'L'：'1'：rest）=L1$：readTrees rest
readTrees（'T'：rest）=Top$：readTrees rest
readTrees[]=Just[]
readTrees=无
Couldn't match type ‘Char -> Maybe Tree’ with ‘Maybe Tree’
Expected type: Char -> Maybe Tree
Actual type: Char -> Char -> Maybe Tree
Probable cause: ‘readtree’ is applied to too few arguments
In the first argument of ‘mapM’, namely ‘readtree’
In the expression: mapM readtree
x $: xs = (x:) <$> xs
readTrees ('L':'1':rest) = L1 $: readTrees rest
readTrees ('T':rest) = Top $: readTrees rest
readTrees [] = Just []
readTrees _ = Nothing
import Control.Monad.Trans.State
readTrees=evalStateT$many readTree>=guard）其中
readTree=StateT uncon>>=\case
'L'->StateT uncon>>=\case
“1”->返回L1
'2'->返回L2
_->空的
'R'->StateT uncon>>=\case
“1”->返回L1
'2'->返回L2
_->空的
‘T’->返回顶部
‘D’->向下返回
_->空的
Couldn't match type ‘Char -> Maybe Tree’ with ‘Maybe Tree’
Expected type: Char -> Maybe Tree
Actual type: Char -> Char -> Maybe Tree
Probable cause: ‘readtree’ is applied to too few arguments
In the first argument of ‘mapM’, namely ‘readtree’
In the expression: mapM readtree
x $: xs = (x:) <$> xs
readTrees ('L':'1':rest) = L1 $: readTrees rest
readTrees ('T':rest) = Top $: readTrees rest
readTrees [] = Just []
readTrees _ = Nothing
import Control.Monad.Trans.State
readTrees=evalStateT$many readTree>=guard）其中
readTree=StateT uncon>>=\case
'L'->StateT uncon>>=\case
“1”->返回L1
'2'->返回L2
_->空的
'R'->StateT uncon>>=\case
“1”->返回L1
'2'->返回L2
_->空的
‘T’->返回顶部
‘D’->向下返回
_->空的
readtreeL1R2DRreadtreeL1R2DR