Buduję analizator składni dla wyrażenia.Przetwarzanie wyrażenia od prawej do lewej
Oto mój przepis gramatyka:
expr ::= term (+ expr | - expr | null)
term ::= expo (* term |/term | null)
expo ::= factor (^ expo | null)
factor ::= (expr) | int
i odpowiedni kod:
expr :: Parser Int
expr = do t <- term
do _ <- symbol "+"
e <- expr
return (t + e)
<|> do _ <- symbol "-"
e <- expr
return (t - e)
<|> return t
term :: Parser Int
term = do ep <- expo
do _ <- symbol "*"
t <- term
return (ep * t)
<|> do _ <- symbol "/"
t <- term
return (ep `div` t)
<|> return ep
expo :: Parser Int
expo = do f <- factor
do _ <- symbol "^"
e <- expo
return (f^e)
<|> return f
factor :: Parser Int
factor = do _ <- symbol "("
e <- expr
_ <- symbol ")"
return e
<|> integer
To działa dobrze dla większości przypadków, ale nie pewne:
$ eval "3*1/3"
od 3 * 1/3
jest analizowana z 3 * (1/3)
(*)
/\
3 (/)
/\
1 3
i
$ eval "4-3-2"
od 4 - 3 - 2
jest analizowana z 4 - (3 - 2)
(-)
/\
4 (-)
/\
3 2
Zdaję sobie sprawę, że wszystko zależy od kierunku analizy (prawostronności).
Czego oczekuję jest (4 - 3) - 2
(-)
/\
(-) 2
/\
4 3
co oznacza chciałbym analizować i interpretować go right-left
left-right
(lub analizować je rekurencyjnie).
Nie mam pojęcia, jak to osiągnąć. Do tej pory przychodzi mi do głowy tylko foldl
.
Czy ktoś może zasugerować, co powinienem zrobić, aby to naprawić?
całego programu:
{-# OPTIONS_GHC -Wall #-}
import Control.Applicative
import Data.Char
newtype Parser a = P (String -> [(a, String)])
parse :: Parser a -> String -> [(a, String)]
parse (P p) inp = p inp
instance Functor Parser where
fmap g p = P (\inp -> case parse p inp of
[] -> []
[(v, out)] -> [(g v, out)]
_ -> undefined)
instance Applicative Parser where
pure v = P (\inp -> [(v, inp)])
pg <*> px = P (\inp -> case parse pg inp of
[] -> []
[(g, out)] -> parse (fmap g px) out
_ -> undefined)
instance Monad Parser where
p >>= f = P (\inp -> case parse p inp of
[] -> []
[(v, out)] -> parse (f v) out
_ -> undefined)
instance Alternative Parser where
empty = P (\_ -> [])
p <|> q = P (\inp -> case parse p inp of
[] -> parse q inp
[(v, out)] -> [(v, out)]
_ -> undefined)
many x = some x <|> pure []
some x = pure (:) <*> x <*> many x
item :: Parser Char
item = P (\inp -> case inp of
[] -> []
(x : xs) -> [(x, xs)])
sat :: (Char -> Bool) -> Parser Char
sat p = do x <- item
if p x
then return x
else empty
digit :: Parser Char
digit = sat isDigit
char :: Char -> Parser Char
char x = sat (== x)
string :: String -> Parser String
string [] = return []
string (x : xs) = do _ <- char x
_ <- string xs
return (x : xs)
space :: Parser()
space = do _ <- many (sat isSpace)
return()
nat :: Parser Int
nat = do xs <- some digit
return (read xs)
int :: Parser Int
int = do _ <- char '-'
n <- nat
return (-n)
<|> nat
token :: Parser a -> Parser a
token p = do _ <- space
v <- p
_ <- space
return v
integer :: Parser Int
integer = token int
symbol :: String -> Parser String
symbol = token . string
expr :: Parser Int
expr = do t <- term
do _ <- symbol "+"
e <- expr
return (t + e)
<|> do _ <- symbol "-"
e <- expr
return (t - e)
<|> return t
term :: Parser Int
term = do ep <- expo
do _ <- symbol "*"
t <- term
return (ep * t)
<|> do _ <- symbol "/"
t <- term
return (ep `div` t)
<|> return ep
expo :: Parser Int
expo = do f <- factor
do _ <- symbol "^"
e <- expo
return (f^e)
<|> return f
factor :: Parser Int
factor = do _ <- symbol "("
e <- expr
_ <- symbol ")"
return e
<|> integer
eval :: String -> Int
eval xs = case (parse expr xs) of
[(n, [])] -> n
[(_, out)] -> error ("Unused input " ++ out)
[] -> error "Invalid input"
_ -> undefined
Spójrz na ['Text.Parsec.Expr'] (https://hackage.haskell.org/package/parsec-3.1.11/docs/Text-Parsec-Expr.html) –