Is there any trick about translating BNF to Parsec program? - parsing

Question

The BNF that match function call chain (like x(y)(z)...):
expr = term T
T = (expr) T
| EMPTY
term = (expr)
| VAR
Translate it to Parsec program that looks so tricky.
term :: Parser Term
term = parens expr <|> var
expr :: Parser Term
expr = do whiteSpace
e <- term
maybeAddSuffix e
where addSuffix e0 = do e1 <- parens expr
maybeAddSuffix $ TermApp e0 e1
maybeAddSuffix e = addSuffix e
<|> return e
Could you list all the design patterns about translating BNF to Parsec program?

Answer 1

The simplest think you could do if your grammar is sizeable is to just use the Alex/Happy combo. It is fairly straightforward to use, accepts the BNF format directly - no human translation needed - and perhaps most importantly, produces blazingly fast parsers/lexers.
If you are dead set on doing it with parsec (or you are doing this as a learning exercise), I find it easier in general to do it in two stages; first lexing, then parsing. Parsec will do both!
First write the appropriate types:
{-# LANGUAGE LambdaCase #-}
import Text.Parsec
import Text.Parsec.Combinator
import Text.Parsec.Prim
import Text.Parsec.Pos
import Text.ParserCombinators.Parsec.Char
import Control.Applicative hiding ((<|>))
import Control.Monad
data Term = App Term Term | Var String deriving (Show, Eq)
data Token = LParen | RParen | Str String deriving (Show, Eq)
type Lexer = Parsec [Char] () -- A lexer accepts a stream of Char
type Parser = Parsec [Token] () -- A parser accepts a stream of Token
Parsing a single token is simple. For simplicity, a variable is 1 or more letters. You can of course change this however you like.
oneToken :: Lexer Token
oneToken = (char '(' >> return LParen) <|>
(char ')' >> return RParen) <|>
(Str <$> many1 letter)
Parsing the entire token stream is just parsing a single token many times, possible separated by whitespace:
lexer :: Lexer [Token]
lexer = spaces >> many1 (oneToken <* spaces)
Note the placement of spaces: this way, white space is accepted at the beginning and end of the string.
Since Parser uses a custom token type, you have to use a custom satisfy function. Fortunately, this is almost identical to the existing satisfy.
satisfy' :: (Token -> Bool) -> Parser Token
satisfy' f = tokenPrim show
(\src _ _ -> incSourceColumn src 1)
(\x -> if f x then Just x else Nothing)
Then we can write parsers for each of the primitive tokens.
lparen = satisfy' $ \case { LParen -> True ; _ -> False }
rparen = satisfy' $ \case { RParen -> True ; _ -> False }
strTok = (\(Str s) -> s) <$> (satisfy' $ \case { Str {} -> True ; _ -> False })
As you may imagine, parens would be useful for our purposes. It is very straightforward to write.
parens :: Parser a -> Parser a
parens = between lparen rparen
Now the interesting parts.
term, expr, var :: Parser Term
term = parens expr <|> var
var = Var <$> strTok
These two should be fairly obvious to you.
Parec contains combinators option and optionMaybe which are useful when you you need to "maybe do something".
expr = do
e0 <- term
option e0 (parens expr >>= \e1 -> return (App e0 e1))
The last line means - try to apply the parser given to option - if it fails, instead return e0.
For testing you can do:
tokAndParse = runParser (lexer <* eof) () "" >=> runParser (expr <* eof) () ""
The eof attached to each parser is to make sure that the entire input is consumed; the string cannot be a member of the grammar if there are extra trailing characters. Note - your example x(y)(z) is not actually in your grammar!
>tokAndParse "x(y)(z)"
Left (line 1, column 5):
unexpected LParen
expecting end of input
But the following is
>tokAndParse "(x(y))(z)"
Right (App (App (Var "x") (Var "y")) (Var "z"))