我想用parsec编写一个解析器。目标是最终能够解析玩具语言。
现在我努力让parsec识别两种不同的可能选项,例如赋值和函数调用。
如何将一个写有 “parseCode” 功能解析如下:
x = 3
y = 4
plus(x,y)
到:
(Assignment "x" "3")
(Assignment "y" "4")
(Invocation "plus" ["x","y"])
感谢
编辑:
**略去了**
编辑2:
我建立了一点您的建议,现在有以下问题 运行parse parseTester "bla" "{plus(3,4)\nmin(2,3)\nx=3\n"
给出了预期的解决方案:Right (Body [Invocation "plus",Invocation "min",Assignment "x" "3"])
。
但在运行功能(几乎)相当于parse parseBody "bla" "{plus(3,4)\nmin(2,3)\nx=3\n}"
导致错误:
Left "bla" (line 4, column 2):
unexpected end of input
expecting white space or "="
我没有看到这个问题。解析器是否突然寻找一个应该在寻找调用的任务?有什么建议么?
代码:
data Body = Body [Statement]
deriving (Show)
data Arguments = Arguments [String]
deriving (Show)
data Statement = Assignment String String
| Invocation String
deriving (Show)
parseBody :: Parser Body
parseBody = do
char '{'
statements <- many1 parseStatement
char '}'
return $ Body statements
parseTester :: Parser Body
parseTester = do
char '{'
x <- many1 parseStatement
return $ Body x
parseStatement :: Parser Statement
parseStatement = do
x <- try parseInvocation <|> parseAssignment <?> "statement"
return x
parseInvocation :: Parser Statement
parseInvocation = do
spaces
name <- many1 (noneOf " (")
spaces
char '('
spaces
bla <- many1 (noneOf ")")
spaces
char ')'
char '\n'
return $ Invocation name
parseAssignment :: Parser Statement
parseAssignment = do
spaces
var <- many1 (noneOf " =")
spaces
char '=' <?> "equal in assignment"
spaces
value <- many1 (noneOf "\n")
char '\n'
spaces
return $ Assignment var value
什么迄今已试过吗? – bheklilr
我已经添加了我现在的代码。基本上,parseBody应该解析语句,但我真的不知道从哪里开始...... – ddccffvv
我看到你的第一个大问题。你试图重新定义内置的'Maybe'类型。只需使用它作为'类型参数=可能字符串',不要自己重新定义它。您还必须更改'parseArguments = return Nothing'才能重新编译它。 – bheklilr