How to parse a bool expression in Haskell - parsing

I am trying to parse a bool expression in Haskell. This line is giving me an error: BoolExpr <$> parseBoolOp <*> (n : ns). This is the error:
• Couldn't match type ‘[]’ with ‘Parser’
Expected type: Parser [Expr]
Actual type: [Expr]
-- define the expression types
data Expr
= BoolExpr BoolOp [Expr]
deriving (Show, Eq)
-- define the type for bool value
data Value
= BoolVal Bool
deriving (Show, Eq)
-- many x = Parser.some x <|> pure []
-- some x = (:) <$> x <*> Parser.many x
kstar :: Alternative f => f a -> f [a]
kstar x = kplus x <|> pure []
kplus :: Alternative f => f a -> f [a]
kplus x = (:) <$> x <*> kstar x
symbol :: String -> Parser String
symbol xs = token (string xs)
-- a bool expression is the operator followed by one or more expressions that we have to parse
-- TODO: add bool expressions
boolExpr :: Parser Expr
boolExpr = do
n <- parseExpr
ns <- kstar (symbol "," >> parseExpr)
BoolExpr <$> parseBoolOp <*> (n : ns)
-- an atom is a literalExpr, which can be an actual literal or some other things
parseAtom :: Parser Expr
parseAtom =
do
literalExpr
-- the main parsing function which alternates between all the options you have
parseExpr :: Parser Expr
parseExpr =
do
parseAtom
<|> parseParens boolExpr
<|> parseParens parseExpr
-- implement parsing bool operations, these are 'and' and 'or'
parseBoolOp :: Parser BoolOp
parseBoolOp =
do symbol "and" >> return And
<|> do symbol "or" >> return Or
The boolExpr is expecting a Parser [Expr] but I am returning only an [Expr]. Is there a way to fix this or do it in another way? When I try pure (n:ns), evalStr "(and true (and false true) true)" returns Left (ParseError "'a' didn't match expected character") instead of Right (BoolVal False)

The expression (n : ns) is a list. Therefore the compiler thinks that the applicative operators <*> and <$> should be used in the context [], while you want Parser instead.
I would guess you need pure (n : ns) instead.

Related

How to re-write a function in Haskell

I am making a Mini lisp language. How can I re-write this using just arrows:
-- a comp expression is the comp operator and the parsing of two expressions
compExpr :: Parser Expr
compExpr = CompExpr <$> parseCompOp <*> parseExpr <*> parseExpr
For reference, here is what the other parsers are:
data CompOp = Eq | Lt deriving (Show, Eq)
-- define the expression types
data Expr
= CompExpr CompOp Expr Expr
deriving (Show, Eq)
parseCompOp :: Parser CompOp
parseCompOp =
do parseKeyword "equal?" >> return Eq
<|> do symbol "<" >> return Lt
-- the main parsing function which alternates between all the options you have
parseExpr :: Parser Expr
parseExpr =
do
parseAtom
<|> parseParens compExpr
<|> parseParens parseExpr
-- an atom is a literalExpr, which can be an actual literal or some other things
parseAtom :: Parser Expr
parseAtom =
do
literalExpr
Here is what I have tried:
compExpr :: Parser Expr
compExpr = do
op <- parseCompOp
expr1 <- parseExpr
expr2 <- parseExpr
return (CompExpr op expr1 expr2)
Since I am not sure how to test this, is this function equivalent to the first function I showed?

Haskell : Operator Parser keeps going to undefined rather than inputs

I'm practicing writing parsers. I'm using Tsodings JSON Parser video as reference. I'm trying to add to it by being able to parse arithmetic of arbitrary length and I have come up with the following AST.
data HVal
= HInteger Integer -- No Support For Floats
| HBool Bool
| HNull
| HString String
| HChar Char
| HList [HVal]
| HObj [(String, HVal)]
deriving (Show, Eq, Read)
data Op -- There's only one operator for the sake of brevity at the moment.
= Add
deriving (Show, Read)
newtype Parser a = Parser {
runParser :: String -> Maybe (String, a)
}
The following functions is my attempt of implementing the operator parser.
ops :: [Char]
ops = ['+']
isOp :: Char -> Bool
isOp c = elem c ops
spanP :: (Char -> Bool) -> Parser String
spanP f = Parser $ \input -> let (token, rest) = span f input
in Just (rest, token)
opLiteral :: Parser String
opLiteral = spanP isOp
sOp :: String -> Op
sOp "+" = Add
sOp _ = undefined
parseOp :: Parser Op
parseOp = sOp <$> (charP '"' *> opLiteral <* charP '"')
The logic above is similar to how strings are parsed therefore my assumption was that the only difference was looking specifically for an operator rather than anything that's not a number between quotation marks. It does seemingly begin to parse correctly but it then gives me the following error:
λ > runParser parseOp "\"+\""
Just ("+\"",*** Exception: Prelude.undefined
CallStack (from HasCallStack):
error, called at libraries/base/GHC/Err.hs:80:14 in base:GHC.Err
undefined, called at /DIRECTORY/parser.hs:110:11 in main:Main
I'm confused as to where the error is occurring. I'm assuming it's to do with sOp mainly due to how the other functions work as intended as the rest of parseOp being a translation of the parseString function:
stringLiteral :: Parser String
stringLiteral = spanP (/= '"')
parseString :: Parser HVal
parseString = HString <$> (charP '"' *> stringLiteral <* charP '"')
The only reason why I have sOp however is that if it was replaced with say Op, I would get the error that the following doesn't exist Op :: String -> Op. When I say this my inclination was that the string coming from the parsed expression would be passed into this function wherein I could return the appropriate operator. This however is incorrect and I'm not sure how to proceed.
charP and Applicative Instance
charP :: Char -> Parser Char
charP x = Parser $ f
where f (y:ys)
| y == x = Just (ys, x)
| otherwise = Nothing
f [] = Nothing
instance Applicative Parser where
pure x = Parser $ \input -> Just (input, x)
(Parser p) <*> (Parser q) = Parser $ \input -> do
(input', f) <- p input
(input', a) <- q input
Just (input', f a)
The implementation of (<*>) is the culprit. You did not use input' in the next call to q, but used input instead. As a result you pass the string to the next parser without "eating" characters. You can fix this with:
instance Applicative Parser where
pure x = Parser $ \input -> Just (input, x)
(Parser p) <*> (Parser q) = Parser $ \input -> do
(input', f) <- p input
(input'', a) <- q input'
Just (input'', f a)
With the updated instance for Applicative, we get:
*Main> runParser parseOp "\"+\""
Just ("",Add)

Haskell : Non-Exhaustive Pattern In Function Prevents Another Function From Executing Even Though Its Not Used

I'm trying to implement car, cdr, and cons functionality into a toy language I'm writing however when I try to execute my car function through main, I get the following error:
./parser "car [1 2 3]"
parser: parser.hs:(48,27)-(55,45): Non-exhaustive patterns in case
The function on lines 48-55 is the following:
parseOp :: Parser HVal
parseOp = (many1 letter <|> string "+" <|> string "-" <|> string "*" <|> string "/" <|> string "%" <|> string "&&" <|> string "||") >>=
(\x -> return $ case x of
"&&" -> Op And
"||" -> Op Or
"+" -> Op Add
"-" -> Op Sub
"*" -> Op Mult
"/" -> Op Div
"%" -> Op Mod)
I'm really unsure why the error message points to this function because it has nothing to do with the list functionality. The car function is working however because I was able to successfully execute it through GHCI. I know my problem is due to parsing but I don't see where it is. The following are the functions that relate to lists. I can't see from them how they are influenced by parseOp.
data HVal = Number Integer
| String String
| Boolean Bool
| List [HVal]
| Op Op
| Expr Op HVal HVal
| Car [HVal]
deriving (Read)
car :: [HVal] -> HVal
car xs = head xs
parseListFunctions :: Parser HVal
parseListFunctions = do
_ <- string "car "
_ <- char '['
x <- parseList
_ <- char ']'
return $ Car [x]
parseExpr :: Parser HVal
parseExpr = parseNumber
<|> parseOp
<|> parseBool
<|> parseListFunctions
<|> do
_ <- char '['
x <- parseList
_ <- char ']'
return x
<|> do
_ <- char '('
x <- parseExpression
_ <- char ')'
return x
eval :: HVal -> HVal
eval val#(Number _) = val
eval val#(String _) = val
eval val#(Boolean _) = val
eval val#(List _) = val -- Look at list eval NOT WORKING
eval val#(Op _) = val
eval (Expr op x y) = eval $ evalExpr (eval x) op (eval y)
eval (Car xs) = eval $ car xs
The removal of many1 letter in parseOp transfers the same error to the following function parseBool:
parseBool :: Parser HVal
parseBool = many1 letter >>= (\x -> return $ case x of
"True" -> Boolean True
"False" -> Boolean False)
You write
parseExpr = ... <|> parseOp <|> ... <|> parseListFunctions <|> ...
and so
car ...
is passed to parseOp first, then parseListFunctions. The parseOp parser succeeds in the
many1 letter
branch, and so in the \x -> return $ case x of ..., x is bound to "car". Because parseOp succeeds (and returns an error value with an embedded, not-yet-evaluated inexhaustive case error!), parseListFunctions is never tried.
You will need to modify your grammar to reduce the ambiguity in it, so that these conflicts where multiple branches may match do not arise.

Removing Left Recursion in a Basic Expression Parser

As an exercise, I'm implementing a parser for an exceedingly simple language defined in Haskell using the following GADT (the real grammar for my project involves many more expressions, but this extract is sufficient for the question):
data Expr a where
I :: Int -> Expr Int
Add :: [Expr Int] -> Expr Int
The parsing functions are as follows:
expr :: Parser (Expr Int)
expr = foldl1 mplus
[ lit
, add
]
lit :: Parser (Expr Int)
lit = I . read <$> some digit
add :: Parser (Expr Int)
add = do
i0 <- expr
is (== '+')
i1 <- expr
is <- many (is (== '+') *> expr)
pure (Add (i0:i1:is))
Due to the left-recursive nature of the expression grammar, when I attempt to parse something as simple as 1+1 using the expr parser, the parser get stuck in an infinite loop.
I've seen examples of how to factor out left recursion across the web using a transformation from something like:
S -> S a | b
Into something like:
S -> b T
T -> a T
But I'm struggling with how to apply this to my parser.
For completeness, here is the code that actually implements the parser:
newtype Parser a = Parser
{ runParser :: String -> [(a, String)]
}
instance Functor Parser where
fmap f (Parser p) = Parser $ \s ->
fmap (\(a, r) -> (f a, r)) (p s)
instance Applicative Parser where
pure a = Parser $ \s -> [(a, s)]
(<*>) (Parser f) (Parser p) = Parser $ \s ->
concat $ fmap (\(f', r) -> fmap (\(a, r') -> (f' a, r')) (p r)) (f >
instance Alternative Parser where
empty = Parser $ \s -> []
(<|>) (Parser a) (Parser b) = Parser $ \s ->
case a s of
(r:rs) -> (r:rs)
[] -> case b s of
(r:rs) -> (r:rs)
[] -> []
instance Monad Parser where
return = pure
(>>=) (Parser a) f = Parser $ \s ->
concat $ fmap (\(r, rs) -> runParser (f r) rs) (a s)
instance MonadPlus Parser where
mzero = empty
mplus (Parser a) (Parser b) = Parser $ \s -> a s ++ b s
char = Parser $ \case (c:cs) -> [(c, cs)]; [] -> []
is p = char >>= \c -> if p c then pure c else empty
digit = is isDigit
Suppose you want to parse non-parenthesized expressions involving literals, addition, and multiplication. You can do this by cutting down the list by precedence. Here's one way to do it in attoparsec, which should be pretty similar to what you'd do with your parser. I'm no parsing expert, so there might be some errors or infelicities.
import Data.Attoparsec.ByteString.Char8
import Control.Applicative
expr :: Parser (Expr Int)
expr = choice [add, mul, lit] <* skipSpace
-- choice is in Data.Attoparsec.Combinators, but is
-- actually a general Alternative operator.
add :: Parser (Expr Int)
add = Add <$> addList
addList :: Parser [Expr Int]
addList = (:) <$> addend <* skipSpace <* char '+' <*> (addList <|> ((:[]) <$> addend))
addend :: Parser (Expr Int)
addend = mul <|> multiplicand
mul :: Parser (Expr Int)
mul = Mul <$> mulList
mulList :: Parser [Expr Int]
mulList = (:) <$> multiplicand <* skipSpace <* char '*' <*> (mulList <|> ((:[]) <$> multiplicand))
multiplicand :: Parser (Expr Int)
multiplicand = lit
lit :: Parser (Expr Int)
lit = I <$> (skipSpace *> decimal)

Writing Parser for S Expressions

I'm trying to write a Parser for S Expressions from Prof. Yorgey's 2013 homework.
newtype Parser a = Parser { runParser :: String -> Maybe (a, String) }
Given the following definitions, presented in the homework:
type Ident = String
-- An "atom" is either an integer value or an identifier.
data Atom = N Integer | I Ident
deriving Show
-- An S-expression is either an atom, or a list of S-expressions.
data SExpr = A Atom
| Comb [SExpr]
deriving Show
I wrote a parser for Parser Atom and Parser SExpr for A Atom.
parseAtom :: Parser Atom
parseAtom = alt n i
where n = (\_ z -> N z) <$> spaces <*> posInt
i = (\ _ z -> I z) <$> spaces <*> ident
parseAAtom :: Parser SExpr
parseAAtom = fmap (\x -> A x) parseAtom
Then, I attempted to write a parser to handle a Parser SExpr for the Comb ... case:
parseComb :: Parser SExpr
parseComb = (\_ _ x _ _ _ -> x) <$> (zeroOrMore spaces) <*> (char '(') <*>
(alt parseAAtom parseComb) <*> (zeroOrMore spaces)
<*> (char ')') <*> (zeroOrMore spaces)
Assuming that parseComb was right, I could simply make usage of oneOrMore for Parser [SExpr].
parseCombElements :: Parser [SExpr]
parseCombElements = oneOrMore parseComb
So, my two last functions compile, but running runParser parseComb "( foo )" never terminates.
What's wrong with my parseComb definition? Please don't give me the whole answer, but rather a hint - for my own learning.
I am very suspicious of zeroOrMore spaces, because spaces is usually a parser which itself parses zero or more spaces. Which means that it can parse the empty string if there aren't any spaces at that point. In particular, the spaces parser always succeeds.
But when you apply zeroOrMore to a parser that always succeeds, the combined parser will never stop - because zeroOrMore only stops trying again once its parser argument fails.
As an aside, Applicative expressions like (\_ _ x _ _ _ -> x) <$> ... <*> ... <*> ...... which only use a single of the subparsers can usually be written more succinctly with the *> and <* combinators:
... *> ... *> x_parser_here <* ... <* ...

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