desugaring arrow语法后如何获得haskell代码?

Ste*_*lla 4 haskell arrows ghc

我目前试图解决我的问题HXT:输入是否可以改变箭头语法?因此,想要在ghc编译器去掉Arrow语法之后看到haskell代码.我怎样才能做到这一点?

我已经尝试-ddump-ds但是有了这个标志,我得到了一个可怕的长代码,因为所有类型都得到了解决.有没有办法看到代码只用箭头语法desugaring?

Zet*_*eta 6

原来箭头项目提供了一个解析器,称为arrowp,这是可在Hackage和转换箭头语法Haskell98:

cabal install arrowp
arrowp --help
arrowp source.hs > desugared.hs
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来源已从您的其他问题中获取.

source.hs

{-# LANGUAGE Arrows #-}
import Text.XML.HXT.Core

data Person = Person { forname :: String, surname :: String } deriving (Show)

parseXml :: IOSArrow XmlTree Person
parseXml = proc x -> do
    forname <- x >- this /> this /> hasName "fn" /> getText
    surname <- x >- this /> this /> hasName "sn" /> getText
    returnA -< Person forname surname

main :: IO ()
main = do
    person <- runX (readString [withValidate no]
                               "<p><fn>John</fn><sn>Smith</sn></p>"
                    >>> parseXml)
    putStrLn $ show person
    return ()
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arrowp source.hs

{-# LINE 2 "source.hs" #-}
module Main (main) where
{-# LINE 2 "source.hs" #-}
import Text.XML.HXT.Core

{-# LINE 4 "source.hs" #-}
data Person = Person{forname :: String, surname :: String}
            deriving Show

{-# LINE 6 "source.hs" #-}
parseXml :: IOSArrow XmlTree Person
{-# LINE 7 "source.hs" #-}
parseXml
  = (arr (\ x -> (x, x)) >>>
       (first (this /> this /> hasName "fn" /> getText) >>>
          arr (\ (forname, x) -> (x, forname)))
         >>>
         (first (this /> this /> hasName "sn" /> getText) >>>
            arr (\ (surname, forname) -> Person forname surname)))

{-# LINE 12 "source.hs" #-}
main :: IO ()
{-# LINE 13 "source.hs" #-}
main
  = do person <- runX
                   (readString [withValidate no] "<p><fn>John</fn><sn>Smith</sn></p>"
                      >>> parseXml)
       putStrLn $ show person
       return ()
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