解析 – 使用Parsec解析正则表达式

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！

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?"