[英]How to parse simple imperative language using Parsec?
我有一種簡單的語言,有以下語法
Expr -> Var | Int | Expr Op Expr
Op -> + | - | * | / | % | == | != | < | > | <= | >= | && | ||
Stmt -> Skip | Var := Expr | Stmt ; Stmt | write Expr | read Expr | while Expr do Stmt | if Expr then Stmt else Stmt
我正在使用Haskell的Parsec庫為該語言編寫簡單的解析器,而我遇到了一些麻煩
當我嘗試解析語句時,請skip ; skip
skip ; skip
我只獲得第一個“ Skip
,但是我想去獲得類似“ Colon Skip Skip
另外,當我嘗試解析分配時,我得到了無限遞歸。 例如,當我嘗試解析x := 1
我的計算機掛斷了很長時間。
這是我的解析器的完整源代碼。 謝謝你的幫助!
module Parser where
import Control.Monad
import Text.Parsec.Language
import Text.ParserCombinators.Parsec
import Text.ParserCombinators.Parsec.Expr
import Text.ParserCombinators.Parsec.Language
import qualified Text.ParserCombinators.Parsec.Token as Token
type Id = String
data Op = Add
| Sub
| Mul
| Div
| Mod
| Eq
| Neq
| Gt
| Geq
| Lt
| Leq
| And
| Or deriving (Eq, Show)
data Expr = Var Id
| Num Integer
| BinOp Op Expr Expr deriving (Eq, Show)
data Stmt = Skip
| Assign Expr Expr
| Colon Stmt Stmt
| Write Expr
| Read Expr
| WhileLoop Expr Stmt
| IfCond Expr Stmt Stmt deriving (Eq, Show)
languageDef =
emptyDef { Token.commentStart = ""
, Token.commentEnd = ""
, Token.commentLine = ""
, Token.nestedComments = False
, Token.caseSensitive = True
, Token.identStart = letter
, Token.identLetter = alphaNum
, Token.reservedNames = [ "skip"
, ";"
, "write"
, "read"
, "while"
, "do"
, "if"
, "then"
, "else"
]
, Token.reservedOpNames = [ "+"
, "-"
, "*"
, "/"
, ":="
, "%"
, "=="
, "!="
, ">"
, ">="
, "<"
, "<="
, "&&"
, "||"
]
}
lexer = Token.makeTokenParser languageDef
identifier = Token.identifier lexer
reserved = Token.reserved lexer
reservedOp = Token.reservedOp lexer
semi = Token.semi lexer
parens = Token.parens lexer
integer = Token.integer lexer
whiteSpace = Token.whiteSpace lexer
ifStmt :: Parser Stmt
ifStmt = do
reserved "if"
cond <- expression
reserved "then"
action1 <- statement
reserved "else"
action2 <- statement
return $ IfCond cond action1 action2
whileStmt :: Parser Stmt
whileStmt = do
reserved "while"
cond <- expression
reserved "do"
action <- statement
return $ WhileLoop cond action
assignStmt :: Parser Stmt
assignStmt = do
var <- expression
reservedOp ":="
expr <- expression
return $ Assign var expr
skipStmt :: Parser Stmt
skipStmt = do
reserved "skip"
return Skip
colonStmt :: Parser Stmt
colonStmt = do
s1 <- statement
reserved ";"
s2 <- statement
return $ Colon s1 s2
readStmt :: Parser Stmt
readStmt = do
reserved "read"
e <- expression
return $ Read e
writeStmt :: Parser Stmt
writeStmt = do
reserved "write"
e <- expression
return $ Write e
statement :: Parser Stmt
statement = colonStmt
<|> assignStmt
<|> writeStmt
<|> readStmt
<|> whileStmt
<|> ifStmt
<|> skipStmt
expression :: Parser Expr
expression = buildExpressionParser operators term
term = fmap Var identifier
<|> fmap Num integer
<|> parens expression
operators = [ [Infix (reservedOp "==" >> return (BinOp Eq)) AssocNone,
Infix (reservedOp "!=" >> return (BinOp Neq)) AssocNone,
Infix (reservedOp ">" >> return (BinOp Gt)) AssocNone,
Infix (reservedOp ">=" >> return (BinOp Geq)) AssocNone,
Infix (reservedOp "<" >> return (BinOp Lt)) AssocNone,
Infix (reservedOp "<=" >> return (BinOp Leq)) AssocNone,
Infix (reservedOp "&&" >> return (BinOp And)) AssocNone,
Infix (reservedOp "||" >> return (BinOp Or)) AssocNone]
, [Infix (reservedOp "*" >> return (BinOp Mul)) AssocLeft,
Infix (reservedOp "/" >> return (BinOp Div)) AssocLeft,
Infix (reservedOp "%" >> return (BinOp Mod)) AssocLeft]
, [Infix (reservedOp "+" >> return (BinOp Add)) AssocLeft,
Infix (reservedOp "-" >> return (BinOp Sub)) AssocLeft]
]
parser :: Parser Stmt
parser = whiteSpace >> statement
parseString :: String -> Stmt
parseString str =
case parse parser "" str of
Left e -> error $ show e
Right r -> r`
這是基於解析器組合器的解析器的一個常見問題: statement
是左遞歸的,因為它的第一個模式是colonStmt
,而colonStmt
要做的第一件事就是嘗試再次解析一個statement
。 眾所周知,解析器組合器在這種情況下不會終止。
從statement
解析器中刪除了colonStmt
模式,其他部分工作正常:
> parseString "if (1 == 1) then skip else skip"
< IfCond (BinOp Eq (Num 1) (Num 1)) Skip Skip
> parseString "x := 1"
< Assign (Var "x") (Num 1)
該解決方案已在此repo中進行了全面描述,沒有許可證文件,因此我真的不知道引用代碼是否安全,通常的想法是在解析任何語句時添加另一層解析器:
statement :: Parser Stmt
statement = do
ss <- sepBy1 statement' (reserved ";")
if length ss == 1
then return $ head ss
else return $ foldr1 Colon ss
statement' :: Parser Stmt
statement' = assignStmt
<|> writeStmt
<|> readStmt
<|> whileStmt
<|> ifStmt
<|> skipStmt
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