我正在尝试通过实现一个小的正则表达式解析器来学习Parsec.在BNF中,我的语法看起来像：
EXP : EXP *
| LIT EXP
| LIT
我试图在Haskell中实现这个：
expr = try star
<|> try litE
<|> lit
litE = do c <- noneOf "*"
rest <- expr
return (c : rest)
lit = do c <- noneOf "*"
return [c]
star = do content <- expr
char '*'
return (content ++ "*")
这里有一些无限循环(例如expr – > star – > expr,不消耗任何标记),这使得解析器永远循环.我不确定如何修复它,因为star的本质是它最终会消耗它的强制令牌.
有什么想法吗？
你应该使用Parsec.Expr.buildExprParser;它非常适合这个目的.您只需描述您的运算符,它们的优先级和关联性,以及如何解析原子,组合器为您构建解析器！
原文链接：https://www.f2er.com/regex/356985.html您可能还希望添加使用parens对术语进行分组的功能,以便您可以将*应用于多个文字.
这是我的尝试(我投入|,和？为了好的衡量标准)：
import Control.Applicative
import Control.Monad
import Text.ParserCombinators.Parsec
import Text.ParserCombinators.Parsec.Expr
data Term = Literal Char
| Sequence [Term]
| Repeat (Int,Maybe Int) Term
| Choice [Term]
deriving ( Show )
term :: Parser Term
term = buildExpressionParser ops atom where
ops = [ [ Postfix (Repeat (0,Nothing) <$char '*'),Postfix (Repeat (1,Nothing) <$char '+'),Postfix (Repeat (0,Just 1) <$char '?')
],[ Infix (return sequence) AssocRight
],[ Infix (choice <$char '|') AssocRight
]
]
atom = msum [ Literal <$> lit,parens term
]
lit = noneOf "*+?|()"
sequence a b = Sequence $(seqTerms a) ++ (seqTerms b)
choice a b = Choice $(choiceTerms a) ++ (choiceTerms b)
parens = between (char '(') (char ')')
seqTerms (Sequence ts) = ts
seqTerms t = [t]
choiceTerms (Choice ts) = ts
choiceTerms t = [t]
main = parseTest term "he(llo)*|wor+ld?"
