Question

我用megaparsec写了一个小解析器：

module App (main) where

import Control.Monad (void)
import Text.Megaparsec
import Text.Megaparsec.String
import qualified Text.Megaparsec.Lexer as L

sc :: Parser ()
sc = L.space (void spaceChar) lineCmnt blockCmnt
  where lineCmnt  = L.skipLineComment "//"
        blockCmnt = L.skipBlockComment "/*" "*/"

symbol :: String -> Parser String
symbol = L.symbol sc

semiParser :: Parser String
semiParser = symbol ";"

main :: IO()
main = do
  input <- getLine
  case parse semiParser input of
    Left val -> putStrLn $ "Failed! "
    Right val -> putStrLn $ "Passed! "
  print "Done"

其中sc和符号以及semiParser来自教程。现在我想使用我的结果，但是我得到了一个类型错误：

App.hs:23:5: error:
    • Couldn't match expected type ‘String
                                    -> Either (ParseError (Token String) Dec) String’
                  with actual type ‘Either t0 t1’
    • In the pattern: Left val
      In a case alternative: Left val -> putStrLn $ "Failed! "
      In a stmt of a 'do' block:
        case parse semiParser input of {
          Left val -> putStrLn $ "Failed! "
          Right val -> putStrLn $ "Passed! " }

App.hs:24:5: error:
    • Couldn't match expected type ‘String
                                    -> Either (ParseError (Token String) Dec) String’
                  with actual type ‘Either t2 t3’
    • In the pattern: Right val
      In a case alternative: Right val -> putStrLn $ "Passed! "
      In a stmt of a 'do' block:
        case parse semiParser input of {
          Left val -> putStrLn $ "Failed! "
          Right val -> putStrLn $ "Passed! " }

所以，我的问题是：如果解析失败，如何匹配String -> Either (ParseError (Token String) Dec) String并获取错误消息或结果？

我想这个问题的一般形式是：模式匹配如何工作？我一般都很困惑如何绑定monad的结果（例如Either在另一个monad中，例如IO（我想我必须模式匹配然后将值提升到上下文中（monad）。

Answer 1

parse :: Stream s Identity t => Parsec s () a -> SourceName -> s -> Either ParseError a 
                                              -- ^^^^^^^^^^ --

你忘记了这个论点。尝试

case parse semiParser "input name here" input of

如何匹配megaparsec结果？

1 个答案: