Question

I have a problem for writing parsec rules for one language
I have next language definition (problematic part)

COMMAND ::= ':' WS LITERAL WS {LITERAL WS}* ';'
LITERAL ::= "[CHAR]*" | [^"\ ][^\ ]*

where WS stands for whitespace and LITERAL is any characters except whitespace or
quoted charactes that can contains white spaces
so, I’ve write next functions:

literal = quotedLiteral <|> many1 (noneOf " ") 
command = do { char ':'
             ; separator
             ; name <- literal
             ; separator
             ; cmds <- endBy literal separator            -- (1)
             ; char ';'                                   -- (2)
             ; return (name, Command cmds)
             }

Problem is that symbol ‘;’ is a valid literal so (1) function parses it, therefore
there is a parsing error, because (2) fails to find ‘;’ character.

Is there any way to overcome this problem:
Either make literal function do not accept ‘;’ as literal or somehow fix (2)?

---

after sclv’s comment I find a solution:

  literal :: Parser Literal
  literal = -- as desired in sclv (changing parserZero to pzero


  command :: Parser TCommand
  command = do { char ':'
            ; separator
            ; name <- literal <?> "no name"
            ; separator
            ; cmds <- sepEndBy (do { try( literal) }) separator
            ; char ';'
            ; return (name, Command cmds)
            }

Answer 1

one (untested) take on solution 1:

literal = quotedLiteral <|> someChars
   where someChars = do 
            res <- many1 (noneOf " \n")
            if (res == ';')
               then parserZero
               else return res