Haskell Megaparsec - Reserved word parsed as identifier









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I'm trying to parse a simple language with lamdba expressions. But runParser expr "lamdbda(x) (return x) returns Right (Var "lamdba") instead of Right (Lambda ["x"] (Return (Var "x")))



My guess is, that I have to add a try somewhere, but I can't figure out where. lambdaExpr parses lamdbas correctly.



Ast.hs



data Expr = Const Integer
| BinOp Op Expr Expr
| Neg Expr
| If Expr Expr Expr
| Var String
| Lambda [String] Stmt
deriving (Show, Eq)

data Op = Multiply
| Divide
| Add
| Subtract
deriving (Show, Eq)

data Stmt = Decl String Expr
| Seq [Stmt]
| Print Expr
| Return Expr
deriving (Show, Eq)


Parser.hs



module Parser where

import Ast

import Control.Monad
import Data.Void
import Control.Monad.Combinators.Expr
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as L

type Parser = Parsec Void String

sc :: Parser ()
sc = L.space space1 lineCmnt blockCmnt
where lineCmnt = L.skipLineComment "--"
blockCmnt = L.skipBlockComment "-" "-"

lexeme :: Parser a -> Parser a
lexeme = L.lexeme sc

symbol :: String -> Parser String
symbol = L.symbol sc

parens :: Parser a -> Parser a
parens = between (symbol "(") (symbol ")")


integer :: Parser Integer
integer = lexeme L.decimal

rword :: String -> Parser ()
rword w = (lexeme . try) (string w *> notFollowedBy alphaNumChar)

rws :: [String] -- list of reserved words
rws = ["if", "then", "else", "let", "print", "lambda", "return"]

identifier :: Parser String
identifier = (lexeme . try) (p >>= check)
where
p = (:) <$> letterChar <*> many alphaNumChar
check x = if x `elem` rws
then fail $ "keyword " ++ show x ++ " cannot be an identifier"
else return x

ifExpr :: Parser Expr
ifExpr = do rword "if"
cond <- expr
rword "then"
thn <- expr
rword "else"
els <- expr
return $ If cond thn els

lambdaExpr :: Parser Expr
lambdaExpr = do rword "lambda"
args <- parens $ sepBy identifier (char ',')
s <- stmt
return $ Lambda args s

expr :: Parser Expr
expr = makeExprParser term operators

term :: Parser Expr
term = parens expr
<|> lambdaExpr
<|> Const <$> integer
<|> Var <$> identifier
<|> ifExpr

operators :: [[Operator Parser Expr]]
operators =
[ [Prefix (Neg <$ symbol "-") ]
, [ InfixL (BinOp Multiply <$ symbol "*")
, InfixL (BinOp Divide <$ symbol "/") ]
, [ InfixL (BinOp Add <$ symbol "+")
, InfixL (BinOp Subtract <$ symbol "-") ]
]

declStmt :: Parser Stmt
declStmt = do rword "let"
name <- identifier
void $ symbol "="
e <- expr
return $ Decl name e

printStmt :: Parser Stmt
printStmt = do rword "print"
e <- expr
return $ Print e

returnStmt :: Parser Stmt
returnStmt = do rword "return"
e <- expr
return $ Return e

stmt :: Parser Stmt
stmt = f <$> sepBy1 stmt' (symbol ";")
where
-- if there's only one stmt return it without using ‘Seq’
f l = if length l == 1 then head l else Seq l

stmt' :: Parser Stmt
stmt' = declStmt
<|> printStmt
<|> returnStmt

runParser :: Parser a -> String -> Either (ParseError (Token String) Void) a
runParser p input = Text.Megaparsec.runParser p "" input









share|improve this question

















  • 1




    Did you misspell "lambda"?
    – danidiaz
    Nov 10 at 13:12






  • 1




    Have you tried using the dbg function to see why it doesn't match? stackage.org/haddock/lts-12.17/megaparsec-6.5.0/… Try adding lambdaExpr = dbg "lambdaExpr" $ do.
    – Icid
    Nov 10 at 13:54










  • I did indeed misspell "lmbda", the debug function helped
    – Philipp Siegmantel
    Nov 10 at 14:05














up vote
2
down vote

favorite
1












I'm trying to parse a simple language with lamdba expressions. But runParser expr "lamdbda(x) (return x) returns Right (Var "lamdba") instead of Right (Lambda ["x"] (Return (Var "x")))



My guess is, that I have to add a try somewhere, but I can't figure out where. lambdaExpr parses lamdbas correctly.



Ast.hs



data Expr = Const Integer
| BinOp Op Expr Expr
| Neg Expr
| If Expr Expr Expr
| Var String
| Lambda [String] Stmt
deriving (Show, Eq)

data Op = Multiply
| Divide
| Add
| Subtract
deriving (Show, Eq)

data Stmt = Decl String Expr
| Seq [Stmt]
| Print Expr
| Return Expr
deriving (Show, Eq)


Parser.hs



module Parser where

import Ast

import Control.Monad
import Data.Void
import Control.Monad.Combinators.Expr
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as L

type Parser = Parsec Void String

sc :: Parser ()
sc = L.space space1 lineCmnt blockCmnt
where lineCmnt = L.skipLineComment "--"
blockCmnt = L.skipBlockComment "-" "-"

lexeme :: Parser a -> Parser a
lexeme = L.lexeme sc

symbol :: String -> Parser String
symbol = L.symbol sc

parens :: Parser a -> Parser a
parens = between (symbol "(") (symbol ")")


integer :: Parser Integer
integer = lexeme L.decimal

rword :: String -> Parser ()
rword w = (lexeme . try) (string w *> notFollowedBy alphaNumChar)

rws :: [String] -- list of reserved words
rws = ["if", "then", "else", "let", "print", "lambda", "return"]

identifier :: Parser String
identifier = (lexeme . try) (p >>= check)
where
p = (:) <$> letterChar <*> many alphaNumChar
check x = if x `elem` rws
then fail $ "keyword " ++ show x ++ " cannot be an identifier"
else return x

ifExpr :: Parser Expr
ifExpr = do rword "if"
cond <- expr
rword "then"
thn <- expr
rword "else"
els <- expr
return $ If cond thn els

lambdaExpr :: Parser Expr
lambdaExpr = do rword "lambda"
args <- parens $ sepBy identifier (char ',')
s <- stmt
return $ Lambda args s

expr :: Parser Expr
expr = makeExprParser term operators

term :: Parser Expr
term = parens expr
<|> lambdaExpr
<|> Const <$> integer
<|> Var <$> identifier
<|> ifExpr

operators :: [[Operator Parser Expr]]
operators =
[ [Prefix (Neg <$ symbol "-") ]
, [ InfixL (BinOp Multiply <$ symbol "*")
, InfixL (BinOp Divide <$ symbol "/") ]
, [ InfixL (BinOp Add <$ symbol "+")
, InfixL (BinOp Subtract <$ symbol "-") ]
]

declStmt :: Parser Stmt
declStmt = do rword "let"
name <- identifier
void $ symbol "="
e <- expr
return $ Decl name e

printStmt :: Parser Stmt
printStmt = do rword "print"
e <- expr
return $ Print e

returnStmt :: Parser Stmt
returnStmt = do rword "return"
e <- expr
return $ Return e

stmt :: Parser Stmt
stmt = f <$> sepBy1 stmt' (symbol ";")
where
-- if there's only one stmt return it without using ‘Seq’
f l = if length l == 1 then head l else Seq l

stmt' :: Parser Stmt
stmt' = declStmt
<|> printStmt
<|> returnStmt

runParser :: Parser a -> String -> Either (ParseError (Token String) Void) a
runParser p input = Text.Megaparsec.runParser p "" input









share|improve this question

















  • 1




    Did you misspell "lambda"?
    – danidiaz
    Nov 10 at 13:12






  • 1




    Have you tried using the dbg function to see why it doesn't match? stackage.org/haddock/lts-12.17/megaparsec-6.5.0/… Try adding lambdaExpr = dbg "lambdaExpr" $ do.
    – Icid
    Nov 10 at 13:54










  • I did indeed misspell "lmbda", the debug function helped
    – Philipp Siegmantel
    Nov 10 at 14:05












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I'm trying to parse a simple language with lamdba expressions. But runParser expr "lamdbda(x) (return x) returns Right (Var "lamdba") instead of Right (Lambda ["x"] (Return (Var "x")))



My guess is, that I have to add a try somewhere, but I can't figure out where. lambdaExpr parses lamdbas correctly.



Ast.hs



data Expr = Const Integer
| BinOp Op Expr Expr
| Neg Expr
| If Expr Expr Expr
| Var String
| Lambda [String] Stmt
deriving (Show, Eq)

data Op = Multiply
| Divide
| Add
| Subtract
deriving (Show, Eq)

data Stmt = Decl String Expr
| Seq [Stmt]
| Print Expr
| Return Expr
deriving (Show, Eq)


Parser.hs



module Parser where

import Ast

import Control.Monad
import Data.Void
import Control.Monad.Combinators.Expr
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as L

type Parser = Parsec Void String

sc :: Parser ()
sc = L.space space1 lineCmnt blockCmnt
where lineCmnt = L.skipLineComment "--"
blockCmnt = L.skipBlockComment "-" "-"

lexeme :: Parser a -> Parser a
lexeme = L.lexeme sc

symbol :: String -> Parser String
symbol = L.symbol sc

parens :: Parser a -> Parser a
parens = between (symbol "(") (symbol ")")


integer :: Parser Integer
integer = lexeme L.decimal

rword :: String -> Parser ()
rword w = (lexeme . try) (string w *> notFollowedBy alphaNumChar)

rws :: [String] -- list of reserved words
rws = ["if", "then", "else", "let", "print", "lambda", "return"]

identifier :: Parser String
identifier = (lexeme . try) (p >>= check)
where
p = (:) <$> letterChar <*> many alphaNumChar
check x = if x `elem` rws
then fail $ "keyword " ++ show x ++ " cannot be an identifier"
else return x

ifExpr :: Parser Expr
ifExpr = do rword "if"
cond <- expr
rword "then"
thn <- expr
rword "else"
els <- expr
return $ If cond thn els

lambdaExpr :: Parser Expr
lambdaExpr = do rword "lambda"
args <- parens $ sepBy identifier (char ',')
s <- stmt
return $ Lambda args s

expr :: Parser Expr
expr = makeExprParser term operators

term :: Parser Expr
term = parens expr
<|> lambdaExpr
<|> Const <$> integer
<|> Var <$> identifier
<|> ifExpr

operators :: [[Operator Parser Expr]]
operators =
[ [Prefix (Neg <$ symbol "-") ]
, [ InfixL (BinOp Multiply <$ symbol "*")
, InfixL (BinOp Divide <$ symbol "/") ]
, [ InfixL (BinOp Add <$ symbol "+")
, InfixL (BinOp Subtract <$ symbol "-") ]
]

declStmt :: Parser Stmt
declStmt = do rword "let"
name <- identifier
void $ symbol "="
e <- expr
return $ Decl name e

printStmt :: Parser Stmt
printStmt = do rword "print"
e <- expr
return $ Print e

returnStmt :: Parser Stmt
returnStmt = do rword "return"
e <- expr
return $ Return e

stmt :: Parser Stmt
stmt = f <$> sepBy1 stmt' (symbol ";")
where
-- if there's only one stmt return it without using ‘Seq’
f l = if length l == 1 then head l else Seq l

stmt' :: Parser Stmt
stmt' = declStmt
<|> printStmt
<|> returnStmt

runParser :: Parser a -> String -> Either (ParseError (Token String) Void) a
runParser p input = Text.Megaparsec.runParser p "" input









share|improve this question













I'm trying to parse a simple language with lamdba expressions. But runParser expr "lamdbda(x) (return x) returns Right (Var "lamdba") instead of Right (Lambda ["x"] (Return (Var "x")))



My guess is, that I have to add a try somewhere, but I can't figure out where. lambdaExpr parses lamdbas correctly.



Ast.hs



data Expr = Const Integer
| BinOp Op Expr Expr
| Neg Expr
| If Expr Expr Expr
| Var String
| Lambda [String] Stmt
deriving (Show, Eq)

data Op = Multiply
| Divide
| Add
| Subtract
deriving (Show, Eq)

data Stmt = Decl String Expr
| Seq [Stmt]
| Print Expr
| Return Expr
deriving (Show, Eq)


Parser.hs



module Parser where

import Ast

import Control.Monad
import Data.Void
import Control.Monad.Combinators.Expr
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as L

type Parser = Parsec Void String

sc :: Parser ()
sc = L.space space1 lineCmnt blockCmnt
where lineCmnt = L.skipLineComment "--"
blockCmnt = L.skipBlockComment "-" "-"

lexeme :: Parser a -> Parser a
lexeme = L.lexeme sc

symbol :: String -> Parser String
symbol = L.symbol sc

parens :: Parser a -> Parser a
parens = between (symbol "(") (symbol ")")


integer :: Parser Integer
integer = lexeme L.decimal

rword :: String -> Parser ()
rword w = (lexeme . try) (string w *> notFollowedBy alphaNumChar)

rws :: [String] -- list of reserved words
rws = ["if", "then", "else", "let", "print", "lambda", "return"]

identifier :: Parser String
identifier = (lexeme . try) (p >>= check)
where
p = (:) <$> letterChar <*> many alphaNumChar
check x = if x `elem` rws
then fail $ "keyword " ++ show x ++ " cannot be an identifier"
else return x

ifExpr :: Parser Expr
ifExpr = do rword "if"
cond <- expr
rword "then"
thn <- expr
rword "else"
els <- expr
return $ If cond thn els

lambdaExpr :: Parser Expr
lambdaExpr = do rword "lambda"
args <- parens $ sepBy identifier (char ',')
s <- stmt
return $ Lambda args s

expr :: Parser Expr
expr = makeExprParser term operators

term :: Parser Expr
term = parens expr
<|> lambdaExpr
<|> Const <$> integer
<|> Var <$> identifier
<|> ifExpr

operators :: [[Operator Parser Expr]]
operators =
[ [Prefix (Neg <$ symbol "-") ]
, [ InfixL (BinOp Multiply <$ symbol "*")
, InfixL (BinOp Divide <$ symbol "/") ]
, [ InfixL (BinOp Add <$ symbol "+")
, InfixL (BinOp Subtract <$ symbol "-") ]
]

declStmt :: Parser Stmt
declStmt = do rword "let"
name <- identifier
void $ symbol "="
e <- expr
return $ Decl name e

printStmt :: Parser Stmt
printStmt = do rword "print"
e <- expr
return $ Print e

returnStmt :: Parser Stmt
returnStmt = do rword "return"
e <- expr
return $ Return e

stmt :: Parser Stmt
stmt = f <$> sepBy1 stmt' (symbol ";")
where
-- if there's only one stmt return it without using ‘Seq’
f l = if length l == 1 then head l else Seq l

stmt' :: Parser Stmt
stmt' = declStmt
<|> printStmt
<|> returnStmt

runParser :: Parser a -> String -> Either (ParseError (Token String) Void) a
runParser p input = Text.Megaparsec.runParser p "" input






haskell megaparsec






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share|improve this question










asked Nov 10 at 12:47









Philipp Siegmantel

53236




53236







  • 1




    Did you misspell "lambda"?
    – danidiaz
    Nov 10 at 13:12






  • 1




    Have you tried using the dbg function to see why it doesn't match? stackage.org/haddock/lts-12.17/megaparsec-6.5.0/… Try adding lambdaExpr = dbg "lambdaExpr" $ do.
    – Icid
    Nov 10 at 13:54










  • I did indeed misspell "lmbda", the debug function helped
    – Philipp Siegmantel
    Nov 10 at 14:05












  • 1




    Did you misspell "lambda"?
    – danidiaz
    Nov 10 at 13:12






  • 1




    Have you tried using the dbg function to see why it doesn't match? stackage.org/haddock/lts-12.17/megaparsec-6.5.0/… Try adding lambdaExpr = dbg "lambdaExpr" $ do.
    – Icid
    Nov 10 at 13:54










  • I did indeed misspell "lmbda", the debug function helped
    – Philipp Siegmantel
    Nov 10 at 14:05







1




1




Did you misspell "lambda"?
– danidiaz
Nov 10 at 13:12




Did you misspell "lambda"?
– danidiaz
Nov 10 at 13:12




1




1




Have you tried using the dbg function to see why it doesn't match? stackage.org/haddock/lts-12.17/megaparsec-6.5.0/… Try adding lambdaExpr = dbg "lambdaExpr" $ do.
– Icid
Nov 10 at 13:54




Have you tried using the dbg function to see why it doesn't match? stackage.org/haddock/lts-12.17/megaparsec-6.5.0/… Try adding lambdaExpr = dbg "lambdaExpr" $ do.
– Icid
Nov 10 at 13:54












I did indeed misspell "lmbda", the debug function helped
– Philipp Siegmantel
Nov 10 at 14:05




I did indeed misspell "lmbda", the debug function helped
– Philipp Siegmantel
Nov 10 at 14:05












1 Answer
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I misspelled "lambda", closing this question.






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        answered Nov 10 at 14:06









        Philipp Siegmantel

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