[Haskell-cafe] [Parsec] Backtracking with try does not work for me?

31 Jul 2006

      I'm trying to write a Parsec parser for a language which authorizes
(this is a simplified example) "a" or "a,b,c" or "a,c" or "a,b". (I
can change the grammar but not the language.)

The first attempt was:

***** CUT HERE ****

import Text.ParserCombinators.Parsec
import System (getArgs)

comma = char ','

minilang = do
       char 'a'
       optional (do {comma ; char 'b'})
       optional (do {comma ; char 'c'})
       eof
       return "OK"

run parser input
        = case (parse parser "" input) of
            Left err -> putStr ("parse error at " ++ (show err) ++ "\n")
            Right x  -> putStr (x ++ "\n")

main = do
  args <- getArgs
  run minilang (head args)

***** CUT HERE ****

Of course, it fails for "a,c":

parse error at (line 1, column 3):
unexpected "c"
expecting "b"

for a reason explained in Parsec's documentation (the parser "optional
(do {comma ; char 'b'})" already consumed the input, do note the
column number).

What puzzles me is that the solution suggested in Parsec's
documentation does not work either:

********* CUT HERE *******************

minilang = do
       char 'a'
       try (optional (do {comma ; char 'b'}))
       optional (do {comma ; char 'c'})
       eof
       return "OK"

********* CUT HERE *******************

parse error at (line 1, column 2):
unexpected "c"
expecting "b"

Apparently, "try" was used (do note that the column number indicates
that there was backtracking) but the parser still fails for
"a,c". Why?

[Haskell-cafe] [Parsec] Backtracking with try does not work for me?

Stephane Bortzmeyer