我想实现一个 read 实例,使我能够读取字符串(例如:“- - 8 - 3 -”)并构造一个包含这些值的列表。
data Value a = Nul | Val a
showsValue :: (Show a) => Value a -> ShowS
showsValue (Val x) = ("Value" ++) . shows x
showsValue (Nul) = ("Nothing 0" ++)
instance Show a => Show (Value a) where
showsPrec _ x = showsValue x
instance Read a => Read (Value a) where
readsPrec _ m = readsMatrix m
readsMatrix :: (Read a) => ReadS (Value a)
readsMatrix ('-':s) = [(Nul, rest) | (' ',rest) <- reads s]
readsMatrix s = [(Val x,rest)| (x,' ':rest) <- reads s]
执行此测试后:
read "- 8 - - 3" :: Value Int
我收到错误*** 异常:Prelude.read:无解析
我在这里做错了什么?
更新
readsMatrix ('-':s) = [(Nul, dropWhile isSpace s)]
readsMatrix s = [(Val x, dropWhile isSpace rest) | (x,rest) <- reads s]
最佳答案
更正值 a 的读取
首先,不用费心删除空格,这一切都可以通过 reads
处理得很好:
readsMatrix :: (Read a) => ReadS (Value a)
readsMatrix ('-':s) = [(Nul, dropWhile (==' ') s)]
readsMatrix s = [(Val x,rest)| (x,rest) <- reads s]
您的读取实例现在适合单个值:
*Main> read "4" :: Value Int
Value4
*Main> read "-" :: Value Int
Nothing 0
更正 [Value a] 的读取
但是您想要读取由空格分隔的列表,因此由于这是非标准行为,您需要编写一个自定义 readList::::ReadS [Value a]
instance Read a => Read (Value a) where
readsPrec _ m = readsMatrix m
readList s = [(map read.words $ s,"")]
那么现在
*Main> read "- 4 2 - 5" :: [Value Int]
[Nothing 0,Value4,Value2,Nothing 0,Value5]
但不幸的是,
*Main> read "- 4 2 \n- 5 4" :: [Value Int]
[Nothing 0,Value4,Value2,Nothing 0,Value5,Value4]
更糟糕的是,
*Main> read "- 4 2 \n- 5 4" :: [[Value Int]]
*** Exception: Prelude.read: no parse
读取矩阵
我认为没有直接的方法来解决这个问题,因为 Read
类中没有 readListOfLists
,所以为什么不创建一个独立的函数
matrix :: Read a => String -> [[Value a]]
matrix = map read.lines
这样
*Main> matrix "3 4 -\n3 - 6\n4 5 -" :: [[Value Int]]
[[Value3,Value4,Nothing 0],[Value3,Nothing 0,Value6],[Value4,Value5,Nothing 0]]
显示不是我会选择的
我认为您的 Value
的 Show
实例有点误导(Nul
上面没有零,Val
未写为Value
)。要么写
data Value a = Nul | Val a deriving Show
使其看起来保持原样或定义它以匹配Read
实例
instance Show a => Show (Value a) where
show Nul = "-"
show (Val a) = show a
关于haskell - 读取实例导致解析错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13553794/