解析函数未定义

Question

我正在遵循以下文件并努力实施。代码中没有提及

parse

或

ord

的实现。我尝试在不实现 Prelude 的情况下做到这一点，但这是我的下一个练习，因此顶部的评论。

我的代码如下，直接从论文中复制：

{- {-# LANGUAGE NoImplicitPrelude #-}

type String :: *
type String = [ Char ]

type Char :: *
data Char = GHC.Types.C# GHC.Prim.Char#
-}

newtype Parser a = Parser (String -> [(a,String)])

item :: Parser Char
item = Parser (\cs -> case cs of
    ""     -> []
    (c:cs) -> [(c,cs)])

class Monad m where
    return :: a -> m a
    (>>=) :: m a -> (a -> m b) -> m b

instance Main.Monad Parser where
      return a = Parser (\cs -> [(a,cs)])
      (>>=) :: Parser a -> (a -> Parser b) -> Parser b
      p >>= f  = Parser (\cs -> concat [parse (f a) cs' |
                                   (a,cs') <- parse p cs])

p :: Parser (Char,Char)
p  = do {c <- item; item; d <- item; Main.return (c,d)}

class Main.Monad m => MonadZero m where
      zero :: m a

class MonadZero m => MonadPlus m where
    (++) :: m a -> m a -> m a

instance MonadZero Parser where
      zero :: Parser a
      zero   = Parser (\cs -> [])

instance MonadPlus Parser where
      p ++ q = Parser (\cs -> parse p cs Main.++ parse q cs)

(+++)  :: Parser a -> Parser a -> Parser a
p +++ q = Parser (\cs -> case parse (p Main.++ q) cs of
                               []     -> []
                               (x:xs) -> [x])

sat  :: (Char -> Bool) -> Parser Char
sat p = do {c <- item; if p c then Main.return c else zero}

char :: Char -> Parser Char
char c = sat (c ==)

string       :: String -> Parser String
string ""     = Main.return ""
string (c:cs) = do {char c; string cs; Main.return (c:cs)}

many   :: Parser a -> Parser [a]
many p  = many1 p +++ Main.return []

many1  :: Parser a -> Parser [a]
many1 p = do {a <- p; as <- many p; Main.return (a:as)}

sepby         :: Parser a -> Parser b -> Parser [a]
p `sepby` sep  = (p `sepby1` sep) +++ Main.return []

sepby1 :: Parser a -> Parser b -> Parser [a] 
p `sepby1` sep = do a <-p
                    as <- many (do {sep; p})
                    Main.return (a:as)

chainl :: Parser a -> Parser (a -> a -> a) -> a -> Parser a 
chainl p op a = (p `chainl1` op) +++ Main.return a

chainl1 :: Parser a -> Parser (a -> a -> a) -> Parser a
p `chainl1` op = do {a <- p; rest a}
                    where
                       rest a = (do f <- op
                                    b <- p
                                    rest (f a b))
                                +++ Main.return a

space :: Parser String
space = many (sat isSpace)

token  :: Parser a -> Parser a
token p = do {a <- p; space; Main.return a}

symb :: String -> Parser String
symb cs = token (string cs)

apply  :: Parser a -> String -> [(a,String)]
apply p = parse (do {space; p})

expr  :: Parser Int
addop :: Parser (Int -> Int -> Int)
mulop :: Parser (Int -> Int -> Int)

expr = term `chainl1` addop
term = factor `chainl1` mulop
factor = digit +++ do {symb "("; n <- expr; symb ")"; Main.return n}
digit = do {x <- token (sat isDigit); Main.return (ord x - ord '0')}

addop = do {symb "+"; Main.return (+)} +++ do {symb "-"; Main.return (-)}
mulop = do {symb "*"; Main.return (*)} +++ do {symb "/"; Main.return (div)}

问题是

parse

函数没有在任何地方定义，当尝试使用 GHCi 进行编译时，出现以下错误：

GHCi, version 9.4.8: https://www.haskell.org/ghc/  :? for help
[1 of 2] Compiling Main             ( MonadicParsingInHaskell.hs, interpreted )

MonadicParsingInHaskell.hs:24:41: error:
    Variable not in scope: parse :: Parser b -> t0 -> [(b, String)]
   |
24 |       p >>= f  = Parser (\cs -> concat [parse (f a) cs' |
   |                                         ^^^^^

MonadicParsingInHaskell.hs:25:47: error:
    Variable not in scope: parse :: Parser a -> String -> [(a, t0)]
   |
25 |                                    (a,cs') <- parse p cs])
   |                                               ^^^^^

MonadicParsingInHaskell.hs:41:31: error:
    Variable not in scope: parse :: Parser a -> String -> [(a, String)]
   |
41 |       p ++ q = Parser (\cs -> parse p cs Main.++ parse q cs)
   |                               ^^^^^

MonadicParsingInHaskell.hs:41:50: error:
    Variable not in scope: parse :: Parser a -> String -> [(a, String)]
   |
41 |       p ++ q = Parser (\cs -> parse p cs Main.++ parse q cs)
   |                                                  ^^^^^

MonadicParsingInHaskell.hs:44:31: error:
    Variable not in scope: parse :: Parser a -> String -> [(a, String)]
   |
44 | p +++ q = Parser (\cs -> case parse (p Main.++ q) cs of
   |                               ^^^^^

MonadicParsingInHaskell.hs:84:19: error:
    Variable not in scope: isSpace :: Char -> Bool
    Suggested fix: Perhaps use ‘space’ (line 84)
   |
84 | space = many (sat isSpace)
   |                   ^^^^^^^

MonadicParsingInHaskell.hs:93:11: error:
    Variable not in scope: parse :: Parser a -> String -> [(a, String)]
   |
93 | apply p = parse (do {space; p})
   |           ^^^^^

MonadicParsingInHaskell.hs:102:29: error:
    Variable not in scope: isDigit :: Char -> Bool
    |
102 | digit = do {x <- token (sat isDigit); Main.return (ord x - ord '0')}
    |                             ^^^^^^^

MonadicParsingInHaskell.hs:102:52: error:
    Variable not in scope: ord :: Char -> b
    Suggested fix:
      Perhaps use one of these:
        ‘or’ (imported from Prelude), ‘odd’ (imported from Prelude)
    |
102 | digit = do {x <- token (sat isDigit); Main.return (ord x - ord '0')}
    |                                                    ^^^

MonadicParsingInHaskell.hs:102:60: error:
    Variable not in scope: ord :: Char -> b
    Suggested fix:
      Perhaps use one of these:
        ‘or’ (imported from Prelude), ‘odd’ (imported from Prelude)
    |
102 | digit = do {x <- token (sat isDigit); Main.return (ord x - ord '0')}
    |                                                            ^^^
Failed, no modules loaded.

我尝试从论文中获取代码进行编译，但没有成功。我认为我需要自己定义

parse

和

ord

函数，但我不确定我是否错误地实现了代码 - 我几乎可以肯定我没有。

Answer 1

parse

实际上是在第 438 页的论文中定义的，无论如何，从 GHC 告诉你它应该有的类型签名来看，它是非常清楚的。

parse :: Parser a -> String -> [(a, String)]
parse (Parser p) = p

ord

和

isSpace

位于

Data.Char

。

其他注意事项：请不要重新实现语言原语（如

Char

）。按照惯例，任何名称中带有

#

的内容都将被视为“低级”和实现细节，甚至不能在没有

MagicHash

扩展名的情况下命名，并且在 >90% 的情况下不应被触及。

NoImplicitPrelude

没有用，除非在极少数情况下您想避免从 Prelude 导入

instances

。否则，简单（且标准！）

import Prelude()

会导致

Prelude

中的所有名称超出范围。您将个人名称添加到导入列表中，而不是重新实现您确实不应该的内容。另外，

*

作为一种已经过时了，应该替换为

Data.Kind.Type