Haskell 失败时解析C链L1
我正在尝试解析来自简单类型lambda演算(F1)的表达式,但我对Parsec有点纠结,我一辈子都不知道如何解决我的问题 我有以下ADT:Haskell 失败时解析C链L1,haskell,parsec,Haskell,Parsec,我正在尝试解析来自简单类型lambda演算(F1)的表达式,但我对Parsec有点纠结,我一辈子都不知道如何解决我的问题 我有以下ADT: newtype LambdaVar = LV Char deriving (Eq, Ord) data Type = TBool | TNat | Arr Type Type deriving Eq data LambdaExpr = Abstr LambdaVar Type LambdaExpr | App LambdaEx
newtype LambdaVar = LV Char
deriving (Eq, Ord)
data Type
= TBool
| TNat
| Arr Type Type
deriving Eq
data LambdaExpr
= Abstr LambdaVar Type LambdaExpr
| App LambdaExpr LambdaExpr
| Var LambdaVar
deriving Eq
newtype TypeContext = TC [(LambdaVar, Type)]
deriving (Eq, Show)
data Expression = Expr TypeContext LambdaExpr Type
deriving (Eq, Show)
使用此解析器:
type ParserT a b = ParsecT String a Identity b
parens :: ParserT a b -> ParserT a b
parens = between (char '(') (char ')')
symbol :: String -> ParserT a String
symbol p = spaces *> string p <* spaces
typeParser :: CharParser () Type
typeParser = arr <|> tbool <|> tnat
where tbool = const TBool <$> string "Bool"
tnat = const TNat <$> string "Nat"
subtyp = parens arr <|> tbool <|> tnat
arr = chainr1 subtyp $ try (symbol "->" *> pure Arr)
lambdaParser :: CharParser () LambdaExpr
lambdaParser = expr
where expr = pApp <|> pAbstr <|> pVar
pVar = Var . LV <$> letter
pAbstr = Abstr <$> (LV <$> (char '\\' *> letter)) <*> (symbol ":" *> typeParser) <*> (char '.' *> expr)
pApp = chainl1 subExpr (char ' ' *> pure App)
subExpr = parens pApp <|> pAbstr <|> pVar
typeContextParser :: CharParser () TypeContext
typeContextParser = TC <$> ((,) <$> (LV <$> letter <* symbol ":") <*> typeParser) `sepBy` try (symbol ",")
expressionParser :: CharParser () Expression
expressionParser = Expr <$> (typeContextParser <* symbol "|-") <*> (lambdaParser <* symbol ":") <*> try typeParser
parse :: String -> Either ParseError Expression
parse = P.parse expressionParser ""
如果我尝试解析它,我会得到一个错误:
unexpected ":"
expecting "(", "\\" or letter
这里发生的事情是,
xy
后面有一个空格,所以解析器假设这是一个应用程序,但是发现了一个它不能作为一个应用程序来解析的:
,但是我不知道如何纠正这个行为。我想我不得不用try
来回溯,但我就是不能让它工作。请包括你的导入-它使你的代码工作更容易
我想我已经通过修改所有的令牌解析器来让您的解析器工作了,同时也消耗了令牌后面的空白
例如,将char x
替换为(char x)纯Arr)
lambdapaser::CharParser()LambdaExpr
lambdapaser=expr
其中expr=pApp-pAbstr-pVar
pVar=Var。LV(词素字母)
pAbstr=absr(LV(lchar'\\'*>letter))(符号“*>typeParser)(lchar'.*>expr)
pApp=链L1子表达式(纯应用程序)
subExpr=parens pApp pAbstr pVar
typeContextParser::CharParser()类型上下文
typeContextParser=TC((,)(感谢您的LV信函!仍然需要一些调整(即chainl1(lexeme subExpr)(纯应用程序)
),但它现在似乎正在工作。
unexpected ":"
expecting "(", "\\" or letter
{-# LANGUAGE NoMonomorphismRestriction, FlexibleContexts #-}
import Text.Parsec
import qualified Text.Parsec as P
import Text.Parsec.Expr
import Text.ParserCombinators.Parsec.Char
import Data.Functor.Identity
newtype LambdaVar = LV Char
deriving (Eq, Ord, Show)
data Type
= TBool
| TNat
| Arr Type Type
deriving (Eq, Show)
data LambdaExpr
= Abstr LambdaVar Type LambdaExpr
| App LambdaExpr LambdaExpr
| Var LambdaVar
deriving (Eq, Show)
newtype TypeContext = TC [(LambdaVar, Type)]
deriving (Eq, Show)
data Expression = Expr TypeContext LambdaExpr Type
deriving (Eq, Show)
type ParserT a b = ParsecT String a Identity b
lexeme p = p <* spaces
lchar = lexeme . char
lstring = lexeme . string
parens :: ParserT a b -> ParserT a b
parens = between (lchar '(') (lchar ')')
symbol :: String -> ParserT a String
symbol p = string p <* spaces
typeParser :: CharParser () Type
typeParser = arr <|> tbool <|> tnat
where tbool = const TBool <$> lstring "Bool"
tnat = const TNat <$> lstring "Nat"
subtyp = parens arr <|> tbool <|> tnat
arr = chainr1 subtyp $ try (symbol "->" *> pure Arr)
lambdaParser :: CharParser () LambdaExpr
lambdaParser = expr
where expr = pApp <|> pAbstr <|> pVar
pVar = Var . LV <$> (lexeme letter)
pAbstr = Abstr <$> (LV <$> (lchar '\\' *> letter)) <*> (symbol ":" *> typeParser) <*> (lchar '.' *> expr)
pApp = chainl1 subExpr (pure App)
subExpr = parens pApp <|> pAbstr <|> pVar
typeContextParser :: CharParser () TypeContext
typeContextParser = TC <$> ((,) <$> (LV <$> letter <* symbol ":") <*> typeParser) `sepBy` try (symbol ",")
expressionParser :: CharParser () Expression
expressionParser = Expr <$> (typeContextParser <* symbol "|-") <*> (lambdaParser <* symbol ":") <*> try typeParser
parseIt :: String -> Either ParseError Expression
parseIt = P.parse expressionParser ""
test1 = parseIt
"|- \\x:Bool -> Nat.\\y:Bool.x y : (Bool -> Nat) -> Bool -> Nat"
-- 1234 56789.123456789 .123456789.
-- 1 2