Is it possible to force backtrack all options? - parsing

Question

I need to parse this syntax for function declaration
foo x = 1
Func "foo" (Ident "x") = 1
foo (x = 1) = 1
Func "foo" (Label "x" 1) = 1
foo x = y = 1
Func "foo" (Ident "x") = (Label "y" 1)
I have written this parser
module SimpleParser where
import Text.Parsec.String (Parser)
import Text.Parsec.Language (emptyDef)
import Text.Parsec
import qualified Text.Parsec.Token as Tok
import Text.Parsec.Char
import Prelude
lexer :: Tok.TokenParser ()
lexer = Tok.makeTokenParser style
where
style = emptyDef {
Tok.identLetter = alphaNum
}
parens :: Parser a -> Parser a
parens = Tok.parens lexer
commaSep :: Parser a -> Parser [a]
commaSep = Tok.commaSep1 lexer
commaSep1 :: Parser a -> Parser [a]
commaSep1 = Tok.commaSep1 lexer
identifier :: Parser String
identifier = Tok.identifier lexer
reservedOp :: String -> Parser ()
reservedOp = Tok.reservedOp lexer
data Expr = IntLit Int | Ident String | Label String Expr | Func String Expr Expr | ExprList [Expr] deriving (Eq, Ord, Show)
integer :: Parser Integer
integer = Tok.integer lexer
litInt :: Parser Expr
litInt = do
n <- integer
return $ IntLit (fromInteger n)
ident :: Parser Expr
ident = Ident <$> identifier
paramLabelItem = litInt <|> paramLabel
paramLabel :: Parser Expr
paramLabel = do
lbl <- try (identifier <* reservedOp "=")
body <- paramLabelItem
return $ Label lbl body
paramItem :: Parser Expr
paramItem = parens paramRecord <|> litInt <|> try paramLabel <|> ident
paramRecord :: Parser Expr
paramRecord = ExprList <$> commaSep1 paramItem
func :: Parser Expr
func = do
name <- identifier
params <- paramRecord
reservedOp "="
body <- paramRecord
return $ (Func name params body)
parseExpr :: String -> Either ParseError Expr
parseExpr s = parse func "" s
I can parse foo (x) = 1 but I cannot parse foo x = 1
parseExpr "foo x = 1"
Left (line 1, column 10):
unexpected end of input
expecting digit, "," or "="
I understand it tries to parse this code like Func "foo" (Label "x" 1) and fails. But after fail why it cannot try to parse it like Func "foo" (Ident "x") = 1
Is there any way to do it?
Also I have tried to swap ident and paramLabel
paramItem = parens paramRecord <|> litInt <|> try paramLabel <|> ident
paramItem = parens paramRecord <|> litInt <|> try ident <|> paramLabel
In this case I can parse foo x = 1 but I cannot parse foo (x = 1) = 2
parseExpr "foo (x = 1) = 2"
Left (line 1, column 8):
unexpected "="
expecting "," or ")"

Answer 1

Here is how I understand how Parsec backtracking works (and doesn't work):
In:
(try a <|> try b <|> c)
if a fails, b will be tried, and if b subsequently fails c will be tried.
However, in:
(try a <|> try b <|> c) >> d
if a succeeds but d fails, Parsec does not go back and try b. Once a succeeded Parsec considers the whole choice as parsed and it moves on to d. It will never go back to try b or c.
This doesn't work either for the same reason:
(try (try a <|> try b <|> c)) >> d
Once a or b succeeds, the whole choice succeeds, and therefor the outer try succeeds. Parsing then moves on to d.
A solution is to distribute d to within the choice:
try (a >> d) <|> try (b >> d) <|> (c >> d)
Now if a succeeds but d fails, b >> d will be tried.