How to re-write a function in Haskell - parsing

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?

Related

How to parse a bool expression in Haskell

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.

Best ADT representation of AST

I have the following grammar for expressions that I'm trying to represent as a Haskell ADT:
Expr = SimpleExpr [OPrelation SimpleExpr]
SimpleExpr = [OPunary] Term {OPadd Term}
Term = Factor {OPmult Factor}
where:
{} means 0 or more
[] means 0 or 1
OPmult, OPadd, OPrelation, OPunary are classes of operators
Note that this grammar does get precedence right.
Here's something I tried:
data Expr = Expr SimpleExpr (Maybe OPrelation) (Maybe SimpleExpr)
data SimpleExpr = SimpleExpr (Maybe OPunary) Term [OPadd] [Term]
data Term = Term Factor [OPmult] [Factor]
which in hindsight I think is awful, especially the [OPadd] [Term] and [OPmult] [Factor] parts. Because, for example, in the parse tree for 1+2+3 it would put [+, +] in one branch and [2, 3] in another, meaning they're decoupled.
What would be a good representation that'll play nice later in the next stages of compilation?
Decomposing { } and [ ] into more data types seems like an overkill
Using lists seems not quite right as it would no longer be a tree (Just a node that's a list)
Maybe for { }. A good idea ?
And finally, I'm assuming after parsing I'll have to pass over the Parse Tree and reduce it to an AST? or should the whole grammar be modified to be less complex? or maybe it's abstract enough?
The AST does not need to be that close to the grammar. The grammar is structured into multiple levels to encode precedence and uses repetition to avoid left-recursion while still being able to correctly handle left-associative operators. The AST does not need to worry about such things.
Instead I'd define the AST like this:
data Expr = BinaryOperation BinaryOperator Expr Expr
| UnaryOperation UnaryOperator Expr
| Literal LiteralValue
| Variable Id
data BinaryOperator = Add | Sub | Mul | Div
data UnaryOperator = Not | Negate
Here's an additional answer that might help you. I don't want to spoil your fun, so here's a very simple example grammar:
-- Expr = Term ['+' Term]
-- Term = Factor ['*' Factor]
-- Factor = number | '(' Expr ')'
-- number = one or more digits
Using a CST
As one approach, we can represent this grammar as a concrete syntax tree (CST):
data Expr = TermE Term | PlusE Term Term deriving (Show)
data Term = FactorT Factor | TimesT Factor Factor deriving (Show)
data Factor = NumberF Int | ParenF Expr deriving (Show)
A Parsec-based parser to turn the concrete syntax into a CST might look like this:
expr :: Parser Expr
expr = do
t1 <- term
(PlusE t1 <$ symbol "+" <*> term)
<|> pure (TermE t1)
term :: Parser Term
term = do
f1 <- factor
(TimesT f1 <$ symbol "*" <*> factor)
<|> pure (FactorT f1)
factor :: Parser Factor
factor = NumberF . read <$> lexeme (many1 (satisfy isDigit))
<|> ParenF <$> between (symbol "(") (symbol ")") expr
with helper functions for whitespace processing:
lexeme :: Parser a -> Parser a
lexeme p = p <* spaces
symbol :: String -> Parser String
symbol = lexeme . string
and main entry point:
parseExpr :: String -> Expr
parseExpr pgm = case parse (spaces *> expr) "(string)" pgm of
Right e -> e
Left err -> error $ show err
after which we can run:
> parseExpr "1+1*(3+4)"
PlusE (FactorT (Number 1)) (TimesT (Number 1) (ParenF (PlusE
(FactorT (Number 3)) (FactorT (Number 4)))))
>
To convert this into the following AST:
data AExpr -- Abstract Expression
= NumberA Int
| PlusA AExpr AExpr
| TimesA AExpr AExpr
we could write:
aexpr :: Expr -> AExpr
aexpr (TermE t) = aterm t
aexpr (PlusE t1 t2) = PlusA (aterm t1) (aterm t2)
aterm :: Term -> AExpr
aterm (FactorT f) = afactor f
aterm (TimesT f1 f2) = TimesA (afactor f1) (afactor f2)
afactor :: Factor -> AExpr
afactor (NumberF n) = NumberA n
afactor (ParenF e) = aexpr e
To interpret the AST, we could use:
interp :: AExpr -> Int
interp (NumberA n) = n
interp (PlusA e1 e2) = interp e1 + interp e2
interp (TimesA e1 e2) = interp e1 * interp e2
and then write:
calc :: String -> Int
calc = interp . aexpr . parseExpr
after which we have a crude little calculator:
> calc "1 + 2 * (6 + 3)"
19
>
Skipping the CST
As an alternative approach, we could replace the parser with one that parses directly into an AST of type AExpr:
expr :: Parser AExpr
expr = do
t1 <- term
(PlusA t1 <$ symbol "+" <*> term)
<|> pure t1
term :: Parser AExpr
term = do
f1 <- factor
(TimesA f1 <$ symbol "*" <*> factor)
<|> pure f1
factor :: Parser AExpr
factor = NumberA . read <$> lexeme (many1 (satisfy isDigit))
<|> between (symbol "(") (symbol ")") expr
You can see how little the structure of these parsers changes. All that's disappeared is the distinction between expressions, terms, and factors at the type level, and constructors like TermE, FactorT, and ParenF whose only purpose is to allow embedding of these types within each other.
In more complex scenarios, the CST and AST parsers might exhibit bigger differences. (For example, in a grammar that allowed 1 + 2 + 3, this might be represented as a single constructor data Expr = ... | PlusE [Term] | ... in the CST but with a nested series of binary PlusA constructors in the same AExpr AST type as above.)
After redefining parseExpr to return an AExpr and dropping the aexpr step from calc, everything else stays the same, and we still have:
> calc "1 + 2 * (6 + 3)"
19
>
Programs for Reference
Here's the full program using an intermediate CST:
-- Calc1.hs, using a CST
{-# OPTIONS_GHC -Wall #-}
module Calc1 where
import Data.Char
import Text.Parsec
import Text.Parsec.String
data Expr = TermE Term | PlusE Term Term deriving (Show)
data Term = FactorT Factor | TimesT Factor Factor deriving (Show)
data Factor = NumberF Int | ParenF Expr deriving (Show)
lexeme :: Parser a -> Parser a
lexeme p = p <* spaces
symbol :: String -> Parser String
symbol = lexeme . string
expr :: Parser Expr
expr = do
t1 <- term
(PlusE t1 <$ symbol "+" <*> term)
<|> pure (TermE t1)
term :: Parser Term
term = do
f1 <- factor
(TimesT f1 <$ symbol "*" <*> factor)
<|> pure (FactorT f1)
factor :: Parser Factor
factor = NumberF . read <$> lexeme (many1 (satisfy isDigit))
<|> ParenF <$> between (symbol "(") (symbol ")") expr
parseExpr :: String -> Expr
parseExpr pgm = case parse (spaces *> expr) "(string)" pgm of
Right e -> e
Left err -> error $ show err
data AExpr -- Abstract Expression
= NumberA Int
| PlusA AExpr AExpr
| TimesA AExpr AExpr
aexpr :: Expr -> AExpr
aexpr (TermE t) = aterm t
aexpr (PlusE t1 t2) = PlusA (aterm t1) (aterm t2)
aterm :: Term -> AExpr
aterm (FactorT f) = afactor f
aterm (TimesT f1 f2) = TimesA (afactor f1) (afactor f2)
afactor :: Factor -> AExpr
afactor (NumberF n) = NumberA n
afactor (ParenF e) = aexpr e
interp :: AExpr -> Int
interp (NumberA n) = n
interp (PlusA e1 e2) = interp e1 + interp e2
interp (TimesA e1 e2) = interp e1 * interp e2
calc :: String -> Int
calc = interp . aexpr . parseExpr
and here's the full program for the more traditional solution that skips an explicit CST representation:
-- Calc2.hs, with direct parsing to AST
{-# OPTIONS_GHC -Wall #-}
module Calc where
import Data.Char
import Text.Parsec
import Text.Parsec.String
lexeme :: Parser a -> Parser a
lexeme p = p <* spaces
symbol :: String -> Parser String
symbol = lexeme . string
expr :: Parser AExpr
expr = do
t1 <- term
(PlusA t1 <$ symbol "+" <*> term)
<|> pure t1
term :: Parser AExpr
term = do
f1 <- factor
(TimesA f1 <$ symbol "*" <*> factor)
<|> pure f1
factor :: Parser AExpr
factor = NumberA . read <$> lexeme (many1 (satisfy isDigit))
<|> between (symbol "(") (symbol ")") expr
parseExpr :: String -> AExpr
parseExpr pgm = case parse (spaces *> expr) "(string)" pgm of
Right e -> e
Left err -> error $ show err
data AExpr -- Abstract Expression
= NumberA Int
| PlusA AExpr AExpr
| TimesA AExpr AExpr
interp :: AExpr -> Int
interp (NumberA n) = n
interp (PlusA e1 e2) = interp e1 + interp e2
interp (TimesA e1 e2) = interp e1 * interp e2
calc :: String -> Int
calc = interp . parseExpr
Okay so Buhr's answer is quite nice. Here's how I did though (no CST) inspired by sepp2k's response:
The AST:
data OP = OPplus | OPminus | OPstar | OPdiv
| OPidiv | OPmod | OPand | OPeq | OPneq
| OPless | OPgreater | OPle | OPge
| OPin | OPor
data Expr =
Relation Expr OP Expr -- > < == >= etc..
| Unary OP Expr -- + -
| Mult Expr OP Expr -- * / div mod and
| Add Expr OP Expr -- + - or
| FactorInt Int | FactorReal Double
| FactorStr String
| FactorTrue | FactorFalse
| FactorNil
| FactorDesig Designator -- identifiers
| FactorNot Expr
| FactorFuncCall FuncCall deriving (Show)
The parsers:
parseExpr :: Parser Expr
parseExpr = (try $ Relation <$>
parseSimpleExpr <*> parseOPrelation <*> parseSimpleExpr)
<|> parseSimpleExpr
parseSimpleExpr :: Parser Expr
parseSimpleExpr = (try simpleAdd)
<|> (try $ Unary <$> parseOPunary <*> simpleAdd)
<|> (try $ Unary <$> parseOPunary <*> parseSimpleExpr)
<|> parseTerm
where simpleAdd = Add <$> parseTerm <*> parseOPadd <*> parseSimpleExpr
parseTerm :: Parser Expr
parseTerm = (try $ Mult <$>
parseFactor <*> parseOPmult <*> parseTerm)
<|> parseFactor
parseFactor :: Parser Expr
parseFactor =
(parseKWnot >> FactorNot <$> parseFactor)
<|> (exactTok "true" >> return FactorTrue)
<|> (exactTok "false" >> return FactorFalse)
<|> (parseNumber)
<|> (FactorStr <$> parseString)
<|> (betweenCharTok '(' ')' parseExpr)
<|> (FactorDesig <$> parseDesignator)
<|> (FactorFuncCall <$> parseFuncCall)
I didn't include basic parsers like parseOPadd as those are what you'd expect and are easy to build.
I still parsed according to the grammar but tweaked it slightly to match my AST.
You could check out the full source which is a compiler for Pascal here.

Convert string to custom data type Haskell [closed]

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I've created a custom datatype Expr
type Name = String
type Domain = [Integer]
data Expr = Val Integer
| Var Name
| Expr :+: Expr
| Expr :-: Expr
| Expr :*: Expr
| Expr :/: Expr
| Expr :%: Expr
deriving Eq
I need to create a parser that creates a datatype Expr from an string without using a parser library in Haskell.
I've created the parser for the initial string that accepts a string of the form "2*a+b" and converts it to the form "Val 2 :*: Var "a" :+: Var "b" " that is accepted by the Expr but this is where I don't know that to do in order to go further. My problem is that I don't know how to create an Expr from a such a string without a parser library.
import Control.Applicative (Alternative (empty, many, some, (<|>)), (<**>))
import Data.Char (isSpace, isDigit)
import Data.Maybe (listToMaybe)
Writing a basic, inefficient parsing library is actually not that hard and can be done in under 50 lines of code. The core type looks like this:
newtype Parser a = Parser (String -> [(a, String)])
parse :: Parser a -> String -> Maybe a
parse (Parser p) s = listToMaybe $ fst <$> p s
This parser partially consumes a string and returns a parsed result a along with the remainder string. But there may be many parsing alternatives, which is why it returns a list of results and remainders.
For working with this type we need a few more utilities. I left the _s for you to implement.
instance Functor Parser where
fmap (Parser p) = _
instance Applicative Parser where
pure a = Parser $ \s -> (a, s) -- Consumes nothing and returns a
Parser pf <*> Parser pa = _ -- Parse pf, then parse pa and apply the result
-- of pf to that of pa.
instance Alternative Parser where
empty = Parser $ \s -> [] -- Matches nothing
Parser p1 <|> Parser p2 = _ -- Matches either p1 or if that fails p2.
satisfy :: (Char -> Bool) -> Parser Char
satisfy = _
space :: Parser ()
space = () <$ satisfy isSpace
spaces :: Parser ()
spaces = () <$ many space
char :: Char -> Parser Char
char c = satisfy (c ==)
-- | Detects the end of file.
eof :: Parser ()
eof = _
-- | Succeeds when at the end of a word, without consuming any input
eow :: Parser ()
eow = _
Now we can go ahead and use this parser like any recursive descent parser:
data Expr = Val Integer
| Expr :*: Expr
| Expr :+: Expr
deriving Show
parseVal :: Parser Expr
parseVal =
char '(' *> parseAdd <* char ')' <|>
Val . read <$> some (satisfy isDigit) <* eow
parseMul :: Parser Expr
parseMul = _
parseAdd :: Parser Expr
parseAdd = _
parseExpr :: Parser Expr
parseExpr = parseAdd <* eof

I am trying to parse a simple expression but it has an infinite loop

I want to parse a language like this
foo = (bar, bar1 = (bar2 = bar4), bar5)
I wrote a simple 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.commaSep lexer
identifier :: Parser String
identifier = Tok.identifier lexer
reservedOp :: String -> Parser ()
reservedOp = Tok.reservedOp lexer
data Expr = Ident String | Label String Expr | ExprList [Expr] deriving (Eq, Ord, Show)
parseExpr :: String -> Either ParseError Expr
parseExpr s = parse expr "" s
expr :: Parser Expr
expr = parens expr
<|> try exprList
<|> ident
ident :: Parser Expr
ident = do
var <- identifier
return $ Ident var
exprLabel :: Parser Expr
exprLabel = do
var <- identifier
reservedOp "="
body <- expr
return $ Label var body
exprList :: Parser Expr
exprList = do
list <- commaSep (try exprLabel <|> expr)
return $ ExprList list
But even with the following simple input it has an infinite loop:
test = parseExpr "foo = bar"
Could someone explain why it doesn't work and how I can fix it?
Thing is, in your code, exprList will loop if it tries to
parse an identifier, that is parse exprList "" "foo" goes into
an infinite loop. This is because it tries to parse it as a list
of either Labels or expressions, and expressions can be lists. Once
it fails to be a exprLabel it tries to see if it can be a expr and so
it calls exprList again.
To fix it you need to make sure that expr checks to see both if it is
a exprLabel or an identifier before it tries exprList. Note that if
all the above fails it will still go into a loop. This is because it doesn't know if this is just the start of a list (or a list of lists of lists of lists...) or not.
To fix this you can make expr only call exprList if it matches a parens, and use exprList as the starting Parser.
expr :: Parser Expr
expr = parens (exprList)
<|> try exprLabel
<|> ident
exprList :: Parser Expr
exprList = do
list <- commaSep expr
return $ ExprList list
And it works like this:
>parse exprList "" "(foo=bar),foo=bar"
Right (ExprList [ExprList [Label "foo" (Ident "bar")],Label "foo" (Ident "bar")])

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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