我正在分离解析器的词法分析和解析阶段。经过一些测试,我意识到当我使用 Parsec 的 Char 令牌以外的其他令牌时,错误消息的帮助不大。
以下是 Parsec 在使用 Char 令牌时的一些错误消息示例:
ghci> P.parseTest (string "asdf" >> spaces >> string "ok") "asdf wrong"
parse error at (line 1, column 7):
unexpected "w"
expecting space or "ok"
ghci> P.parseTest (choice [string "ok", string "nop"]) "wrong"
parse error at (line 1, column 1):
unexpected "w"
expecting "ok" or "nop"
因此,字符串解析器会在发现意外字符串时显示预期的字符串,而选择解析器会显示备选字符串。
但是当我对我的标记使用相同的组合器时:
ghci> Parser.parseTest ((tok $ Ide "asdf") >> (tok $ Ide "ok")) "asdf "
parse error at "test" (line 1, column 1):
unexpected end of input
在这种情况下,它不会打印预期的内容。
ghci> Parser.parseTest (choice [tok $ Ide "ok", tok $ Ide "nop"]) "asdf "
parse error at (line 1, column 1):
unexpected (Ide "asdf","test" (line 1, column 1))
当我使用时choice
,它不会打印替代品。
我希望这种行为与组合函数有关,而不是与令牌有关,但似乎我错了。我怎样才能解决这个问题?
这是完整的词法分析器+解析器代码:
词法分析器:
module Lexer
( Token(..)
, TokenPos(..)
, tokenize
) where
import Text.ParserCombinators.Parsec hiding (token, tokens)
import Control.Applicative ((<*), (*>), (<$>), (<*>))
data Token = Ide String
| Number String
| Bool String
| LBrack
| RBrack
| LBrace
| RBrace
| Keyword String
deriving (Show, Eq)
type TokenPos = (Token, SourcePos)
ide :: Parser TokenPos
ide = do
pos <- getPosition
fc <- oneOf firstChar
r <- optionMaybe (many $ oneOf rest)
spaces
return $ flip (,) pos $ case r of
Nothing -> Ide [fc]
Just s -> Ide $ [fc] ++ s
where firstChar = ['A'..'Z'] ++ ['a'..'z'] ++ "_"
rest = firstChar ++ ['0'..'9']
parsePos p = (,) <$> p <*> getPosition
lbrack = parsePos $ char '[' >> return LBrack
rbrack = parsePos $ char ']' >> return RBrack
lbrace = parsePos $ char '{' >> return LBrace
rbrace = parsePos $ char '}' >> return RBrace
token = choice
[ ide
, lbrack
, rbrack
, lbrace
, rbrace
]
tokens = spaces *> many (token <* spaces)
tokenize :: SourceName -> String -> Either ParseError [TokenPos]
tokenize = runParser tokens ()
解析器:
module Parser where
import Text.Parsec as P
import Control.Monad.Identity
import Lexer
parseTest :: Show a => Parsec [TokenPos] () a -> String -> IO ()
parseTest p s =
case tokenize "test" s of
Left e -> putStrLn $ show e
Right ts' -> P.parseTest p ts'
tok :: Token -> ParsecT [TokenPos] () Identity Token
tok t = token show snd test
where test (t', _) = case t == t' of
False -> Nothing
True -> Just t
解决方案:
好的,在 fp4me 的回答和更仔细地阅读 Parsec 的 Char 源之后,我得到了这个:
{-# LANGUAGE FlexibleContexts #-}
module Parser where
import Text.Parsec as P
import Control.Monad.Identity
import Lexer
parseTest :: Show a => Parsec [TokenPos] () a -> String -> IO ()
parseTest p s =
case tokenize "test" s of
Left e -> putStrLn $ show e
Right ts' -> P.parseTest p ts'
type Parser a = Parsec [TokenPos] () a
advance :: SourcePos -> t -> [TokenPos] -> SourcePos
advance _ _ ((_, pos) : _) = pos
advance pos _ [] = pos
satisfy :: (TokenPos -> Bool) -> Parser Token
satisfy f = tokenPrim show
advance
(\c -> if f c then Just (fst c) else Nothing)
tok :: Token -> ParsecT [TokenPos] () Identity Token
tok t = (Parser.satisfy $ (== t) . fst) <?> show t
现在我收到相同的错误消息:
ghci> Parser.parseTest (choice [tok $ Ide "ok", tok $ Ide "nop"]) "asdf"
解析错误 (第 1 行,第 1 列):
意外 (Ide "asdf","test" (第 1 行, 列 3))
期待 Ide "ok" 或 Ide "nop"