Question

我正在编写简单的sitemap.xml搜寻器。代码如下。我的问题是为什么main末尾的代码不会打印任何内容。我怀疑它是因为哈斯克尔的懒惰但不知道如何在这里处理它：

import Network.HTTP.Conduit
import qualified Data.ByteString.Lazy as L
import Text.XML.Light
import Control.Monad.Trans (liftIO)
import Control.Monad
import Data.String.Utils
import Control.Exception

download :: Manager -> Request -> IO (Either HttpException L.ByteString)
download manager req = do
  try $
    fmap responseBody (httpLbs req manager)

downloadUrl :: Manager -> String -> IO (Either HttpException L.ByteString)
downloadUrl manager url = do
  request <- parseUrl url
  download manager request

getPages :: Manager -> [String] -> IO [Either HttpException L.ByteString]
getPages manager urls =
  sequence $ map (downloadUrl manager) urls

main = withManager $ \ manager -> do
  -- I know simpleHttp is bad here
  mapSource <- liftIO $ simpleHttp "http://example.com/sitemap.xml"

  let elements = (parseXMLDoc mapSource) >>= Just . findElements (mapElement "loc")
      Just urls = liftM (map $ (replace "/#!" "?_escaped_fragment_=") . strContent) elements
      mapElement name = QName name (Just "http://www.sitemaps.org/schemas/sitemap/0.9") Nothing

  return $
    getPages manager urls >>= \ pages -> do
      print "evaluate me!"
      sequence $ map print pages

Answer 1

用return（http://hackage.haskell.org/package/resourcet-1.1.1/docs/Control-Monad-Trans-Resource.html#v:runResourceT）替换您的上一个runResourceT。正如它的类型所示，它会将ResourceT转变为IO行动。

Answer 2

你遇到了我在这里描述的同样的问题，至少就错误的代码而言，它实际上应该给出类型错误：Why is the type of "Main.main", "IO ()" and not "IO a"?。这就是为什么你应该总是明确地给main类型签名main :: IO ()。

要解决此问题，您需要将return替换为lift（请参阅http://hackage.haskell.org/package/transformers/docs/Control-Monad-Trans-Class.html#v:lift），并将sequence $ map ...替换为mapM_。 mapM_ f相当于sequence_ . map f。

Haskell不评估块

2 个答案: