Fast parsing of string that allows escaped characters?
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E

1

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I'm trying to parse a string that can contain escaped characters, here's an example:

import qualified Data.Text as T

exampleParser :: Parser T.Text
exampleParser = T.pack <$> many (char '\\' *> escaped <|> anyChar)
  where escaped = satisfy (\c -> c `elem` ['\\', '"', '[', ']'])

The parser above creates a String and then packs it into Text. Is there any way to parse a string with escapes like the above using the functions for efficient string handling that attoparsec provides? Like string, scan, runScanner, takeWhile, ...

Parsing something like "one \"two\" \[three\]" would produce one "two" [three].

Update:

Thanks to @epsilonhalbe I was able to come out with a generalized solution perfect for my needs; note that the following function doesn't look for matching escaped characters like [..], "..", (..), etc; and also, if it finds an escaped character that is not valid it treats \ as a literal character.

takeEscapedWhile :: (Char -> Bool) -> (Char -> Bool) -> Parser Text
takeEscapedWhile isEscapable while = do
  x <- normal
  xs <- many escaped
  return $ T.concat (x:xs)
  where normal = Atto.takeWhile (\c -> c /= '\\' && while c)
        escaped = do
          x <- (char '\\' *> satisfy isEscapable) <|> char '\\'
          xs <- normal
          return $ T.cons x xs
Eurypterid answered 9/2, 2016 at 19:42 Comment(0)
G
6

It is possible writing some escaping code, attoparsec and text - altogether it is pretty straightforward - seeing you have already worked with parsers

import Data.Attoparsec.Text as AT
import qualified Data.Text as T
import Data.Text (Text)

escaped, quoted, brackted :: Parser Text
normal =  AT.takeWhile (/= '\\')
escaped = do r <- normal
             rs <- many escaped'
             return $ T.concat $ r:rs
  where escaped' = do r1 <- normal
                      r2 <- quoted <|> brackted
                      return $ r1 <> r2

quoted = do string "\\\""
            res <- normal
            string "\\\""
            return $ "\""<>res <>"\""

brackted = do string "\\["
              res <- normal
              string "\\]"
              return $ "["<>res<>"]"

then you can use it to parse the following test cases

Prelude >: MyModule
Prelude MyModule> import Data.Attoparsec.Text as AT
Prelude MyModule AT> import Data.Text.IO as TIO
Prelude MyModule AT TIO>:set -XOverloadedStrings
Prelude MyModule AT TIO> TIO.putStrLn $ parseOnly escaped "test"
test
Prelude MyModule AT TIO> TIO.putStrLn $ parseOnly escaped "\\\"test\\\""
"test"
Prelude MyModule AT TIO> TIO.putStrLn $ parseOnly escaped "\\[test\\]"
[test]
Prelude MyModule AT TIO> TIO.putStrLn $ parseOnly escaped "test \\\"test\\\" \\[test\\]"
test "test" [test]

note you have to escape the escapes - that's why you see \\\" instead of \"

Also if you just parse it will print the Text values escaped, like

Right "test \"text\" [test]"

for the last example.

If you parse a file you write simpley escaped text in the file.

test.txt

I \[like\] \"Haskell\"

then you can

Prelude MyModule AT TIO> file <- TIO.readFile "test.txt" 
Prelude MyModule AT TIO> TIO.putStrLn $ parseOnly escaped file
I [like] "Haskell"
Gingham answered 9/2, 2016 at 21:48 Comment(1)
The code required makes you think if the trade of performance over simplicity is worth itEurypterid

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