Question

我正在为Yesod应用程序进行OAuth2身份验证，我遇到了一个我真的不太了解的类型错误。目前代码已被破解，我有几个:: IO ()和undefined被抛出来帮助我隔离类型错误，但相关代码是：

getAccessToken :: Manager -> OAuth2 -> ExchangeToken -> IO (OAuth2Result Errors OAuth2Token)
getAccessToken manager oa code = do

  let (uri, defaultBody) = accessTokenUrl oa code
  let body = defaultBody <> [ ("client_id", TE.encodeUtf8 . oauthClientId $ oa )
                            , ("client_secret", TE.encodeUtf8 . oauthClientSecret $ oa)
                            , ("resource", TE.encodeUtf8 . oauthClientId $ oa)
                            ]

  response <- performOAuth2PostRequest manager oa uri body

  return undefined

performOAuth2PostRequest :: Manager -> OAuth2 -> URI -> PostBody -> IO (Response ByteString)
performOAuth2PostRequest manager oa uri body  = do
  defaultReq <- uriToRequest uri

  let addBasicAuth = applyBasicAuth (TE.encodeUtf8 . oauthClientId $ oa)
                                    (TE.encodeUtf8 . oauthClientSecret $ oa)

  let req = (addBasicAuth . updateRequestHeaders Nothing) defaultReq

  (httpLbs (urlEncodedBody body req) manager) :: IO (Response ByteString)

请注意，我使用httpLbs (urlEnc...) manager扩展程序专门将IO (Response ByteString)操作的类型设置为ScopedTypeVariables。此外，该行代码应该是一个IO操作，因为它是在IO操作的顶层执行的。

事实上，我参加了一次GHCi会议并做了：

Network.OAuth.OAuth2.HttpClient Network.OAuth.OAuth2.Internal 
Network.HTTP.Conduit Data.Functor Prelude> :t httpLbs
httpLbs
  :: Control.Monad.IO.Class.MonadIO m =>
     Request
     -> Manager -> m (Response Data.ByteString.Lazy.Internal.ByteString)

这证实了我理解httpLbs应该产生MonadIO m => m (Response ByteString)。

但这是我得到的错误：

• Couldn't match type ‘Response
                         Data.ByteString.Lazy.Internal.ByteString’
                 with ‘IO (Response ByteString)’
  Expected type: Manager -> IO (Response ByteString)
    Actual type: Manager
                 -> Response Data.ByteString.Lazy.Internal.ByteString
• The function ‘httpLbs’ is applied to two arguments,
  its type is ‘Request
               -> m1 (Response Data.ByteString.Lazy.Internal.ByteString)’,
  it is specialized to ‘Request
                        -> Manager -> Response Data.ByteString.Lazy.Internal.ByteString’

为什么GHC将m专门化为Response而不是IO？我该如何解决？

Answer 1

您还没有包含导入语句，因此很难对其进行调试。我最好的猜测是你导入了Network.HTTP.Simple，它提供了不需要显式Manager参数的函数。我从提供预期类型的错误消息中猜到了这一点：

Request -> m1 (Response Data.ByteString.Lazy.Internal.ByteString)

解决方案：更改导入，或删除Manager参数。

使用HTTP-Conduit时Haskell类型错误

1 个答案: