Question

我正在尝试为基于缩进的语法编写词法分析器，但是我在匹配缩进时遇到了问题。

这是我的代码：

{
module Lexer ( main ) where

import System.IO.Unsafe
}


%wrapper "monadUserState"

$whitespace = [\ \t\b]
$digit      = 0-9                                            -- digits
$alpha      = [A-Za-z]
$letter     = [a-zA-Z]                                       -- alphabetic characters
$ident      = [$letter $digit _]                             -- identifier character
$indent     = [\ \t]

@number     = [$digit]+
@identifier = $alpha($alpha|_|$digit)*

error:-

@identifier { mkL LVarId }

\n $whitespace* \n { skip }
\n $whitespace*    { setIndent }
$whitespace+       { skip }

{

data Lexeme = Lexeme AlexPosn LexemeClass (Maybe String)

instance Show Lexeme where
    show (Lexeme _ LEOF _)   = "  Lexeme EOF"
    show (Lexeme p cl  mbs) = " Lexeme class=" ++ show cl ++ showap p ++ showst mbs
      where
        showap pp = " posn=" ++ showPosn pp
        showst Nothing  = ""
        showst (Just s) = " string=" ++ show s

instance Eq Lexeme where
    (Lexeme _ cls1 _) == (Lexeme _ cls2 _) = cls1 == cls2

showPosn :: AlexPosn -> String
showPosn (AlexPn _ line col) = show line ++ ':': show col

tokPosn :: Lexeme -> AlexPosn
tokPosn (Lexeme p _ _) = p

data LexemeClass
    = LVarId
    | LTIndent Int
    | LTDedent Int
    | LIndent
    | LDedent
    | LEOF
    deriving (Show, Eq)

mkL :: LexemeClass -> AlexInput -> Int -> Alex Lexeme
mkL c (p, _, _, str) len = return (Lexeme p c (Just (take len str)))

data AlexUserState = AlexUserState { indent :: Int }

alexInitUserState :: AlexUserState
alexInitUserState = AlexUserState 0

type Action = AlexInput -> Int -> Alex Lexeme

getLexerIndentLevel :: Alex Int
getLexerIndentLevel = Alex $ \s@AlexState{alex_ust=ust} -> Right (s, indent ust)

setLexerIndentLevel :: Int -> Alex ()
setLexerIndentLevel i = Alex $ \s@AlexState{alex_ust=ust} -> Right (s{alex_ust=(AlexUserState i)}, ())

setIndent :: Action
setIndent input@(p, _, _, str) i = do
    --let !x = unsafePerformIO $ putStrLn $ "|matched string: " ++ str ++ "|"
    lastIndent <- getLexerIndentLevel
    currIndent <- countIndent (drop 1 str) 0 -- first char is always \n
    if (lastIndent < currIndent) then
        do setLexerIndentLevel currIndent
           mkL (LTIndent (currIndent - lastIndent)) input i
    else if (lastIndent > currIndent) then
        do setLexerIndentLevel currIndent
           mkL (LTDedent (lastIndent - currIndent)) input i
    else alexMonadScan
  where
    countIndent str total
        | take 1 str == "\t" = do skip input 1
                                  countIndent (drop 1 str) (total+1)
        | take 4 str == "    " = do skip input 4
                                    countIndent (drop 4 str) (total+1)
        | otherwise = return total

alexEOF :: Alex Lexeme
alexEOF = return (Lexeme undefined LEOF Nothing)

scanner :: String -> Either String [Lexeme]
scanner str =
    let loop = do
        tok@(Lexeme _ cl _) <- alexMonadScan
        if (cl == LEOF)
            then return [tok]
            else do toks <- loop
                    return (tok:toks)
    in runAlex str loop

addIndentations :: [Lexeme] -> [Lexeme]
addIndentations (lex@(Lexeme pos (LTIndent c) _):ls) =
    concat [iter lex c, addIndentations ls]
  where iter lex c = if c == 0 then []
                     else (Lexeme pos LIndent Nothing):(iter lex (c-1))
addIndentations (lex@(Lexeme pos (LTDedent c) _):ls) =
    concat [iter lex c, addIndentations ls]
  where iter lex c = if c == 0 then []
                     else (Lexeme pos LDedent Nothing):(iter lex (c-1))
addIndentations (l:ls) = l:(addIndentations ls)
addIndentations [] = []


main = do
    s <- getContents
    return ()
    print $ fmap addIndentations (scanner s)

}

问题是在行\n $whitespace* { setIndent }中，正则表达式匹配错误的字符串并使用此错误的字符串调用setIndent。出于调试目的，我在unsafePerformIO函数中添加了setIndent，这是程序的示例运行：

begin       
        first indent
|matched string: 
        first indent
                second indent
                second indent
dedent
dedent
|
|matched string: 
                second indent
dedent
|
|matched string: 
dedent
|
|matched string: 
|
Right [ Lexeme class=LVarId posn=1:1 string="begin", Lexeme class=LIndent posn=1:6, Lexeme class=LVarId posn=2:15 string="indent", Lexeme class=LIndent posn=2:21, Lexeme class=LDedent posn=3:30, Lexeme class=LDedent posn=3:30, Lexeme class=LVarId posn=4:1 string="dedent",  Lexeme EOF]

因此调用setIndent不仅仅是空格。并且在它返回缩进的lexeme之后，字符串的其他部分被省略。

这是Alex的错误吗？或者我做错了什么？

Answer 1

所以我没有详细分析你的代码，但我注意到了这一点：

setIndent :: Action
setIndent input@(p, _, _, str) i = do
    --let !x = unsafePerformIO $ putStrLn $ "|matched string: " ++ str ++ "|"

请注意，str是输入的其余部分，而不仅仅是当前令牌。要获取当前令牌，您需要take i str。也许这给你的印象是令牌匹配的输入比实际更多。

我们当然会在GHC自己的词法分析器中处理缩进，所以你可能想要look there for ideas（尽管你可能认为它相当大而复杂）。

Haskell Alex - 正则表达式匹配错误的字符串？

1 个答案: