全部，
我正在尝试使用 parsec 编写解析器。目标是最终能够解析一种玩具语言。
现在我正在努力让 parsec 识别两种不同的可能选项，例如赋值和函数调用。
如何编写一个“parseCode”函数来解析以下内容：
x = 3
y = 4
plus(x,y)
进入：
(Assignment "x" "3")
(Assignment "y" "4")
(Invocation "plus" ["x","y"])
谢谢
编辑：
**为简洁起见省略**
编辑2：
我根据您的建议进行了一些构建，现在遇到以下问题 Runningparse 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"]).
但是运行功能上（几乎）等效parse parseBody "bla" "{plus(3,4)\nmin(2,3)\nx=3\n}"的结果会导致错误：
Left "bla" (line 4, column 2):
unexpected end of input
expecting white space or "="
我没有看到问题。解析器是否突然在寻找它应该寻找调用的分配？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