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I want to create a parser combinator, which will collect all lines below current place, which indentation levels will be greater or equal some i. I think the idea is simple:

Consume a line - if its indentation is:

  • ok -> do it for next lines
  • wrong -> fail

Lets consider following code:

import qualified Text.ParserCombinators.UU as UU
import           Text.ParserCombinators.UU hiding(parse)
import           Text.ParserCombinators.UU.BasicInstances hiding (Parser)

-- end of line
pEOL   = pSym '\n'

pSpace = pSym ' '
pTab   = pSym '\t'

indentOf s = case s of
    ' '  -> 1
    '\t' -> 4

-- return the indentation level (number of spaces on the beginning of the line)
pIndent = (+) <$> (indentOf <$> (pSpace <|> pTab)) <*> pIndent `opt` 0

-- returns tuple of (indentation level, result of parsing the second argument)
pIndentLine p = (,) <$> pIndent <*> p <* pEOL

-- SHOULD collect all lines below witch indentations greater or equal i
myParse p i = do
    (lind, expr) <- pIndentLine p
    if lind < i
        then pFail
        else do
            rest <- myParse p i `opt` []
            return $ expr:rest

-- sample inputs
s1 = " a\
   \\n a\
   \\n"

s2 = " a\
   \\na\
   \\n"

-- execution
pProgram = myParse (pSym 'a') 1 

parse p s = UU.parse ( (,) <$> p <*> pEnd) (createStr (LineColPos 0 0 0) s)

main :: IO ()
main = do 
    print $ parse pProgram s1
    print $ parse pProgram s2
    return ()

Which gives following output:

("aa",[])
Test.hs: no correcting alternative found

The result for s1 is correct. The result for s2 should consume first "a" and stop consuming. Where this error comes from?

4

1 回答 1

1

您正在构建的解析器将始终尝试继续；如有必要，输入将被丢弃或添加。然而pFail是个死胡同。它充当 的单位元素<|>。

但是，在您的解析器中，如果输入不符合解析器识别的语言，则没有其他替代方案。在您的规范中，您说您希望解析器在输入 s2 上失败。现在它失败并显示一条消息说失败，你很惊讶。

也许您不希望它失败，但您想停止接受进一步的输入？在这种情况下，替换pFail为return [].

注意文字：

do
    rest <- myParse p i `opt` []
    return $ expr:rest

可以替换为(expr:) <$> (myParse p i `opt` [])

解决问题的自然方法可能是

pIndented p = do i <- pGetIndent
             (:) <$> p <* pEOL  <*> pMany (pToken (take i (repeat ' ')) *> p <* pEOL)

pIndent = length <$> pMany (pSym ' ')
于 2013-08-15T08:13:40.947 回答