Question

我正在学习Haskell，编译器给了我一个我不太明白的错误。它说期望的和实际的类型不匹配，但是对我来说它们看起来是相同的。谁能帮助我了解此错误试图说明什么？

/app/app/Main.hs:75:11: error:
    • Couldn't match type ‘a’ with ‘a1’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          app :: forall a.
                 (ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a
                  -> IO a)
                 -> IO Middleware
        at /app/app/Main.hs:68:1-99
      ‘a1’ is a rigid type variable bound by
        a type expected by the context:
          forall a1.
          ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a1
          -> IO a1
        at /app/app/Main.hs:75:3-12
      Expected type: ReaderT
                       (W.Options -> String -> IO (Response BL.ByteString)) IO a1
                     -> IO a1
        Actual type: ReaderT
                       (W.Options -> String -> IO (Response BL.ByteString)) IO a
                     -> IO a
    • In the first argument of ‘spockT’, namely ‘(r)’
      In the expression: spockT (r)
      In a stmt of a 'do' block: spockT (r) $ routes apiKey template
    • Relevant bindings include
        r :: ReaderT
               (W.Options -> String -> IO (Response BL.ByteString)) IO a
             -> IO a
          (bound at /app/app/Main.hs:69:5)
        app :: (ReaderT
                  (W.Options -> String -> IO (Response BL.ByteString)) IO a
                -> IO a)
               -> IO Middleware
          (bound at /app/app/Main.hs:69:1)
   |
75 |   spockT (r) $ routes apiKey template
   |           ^

让我更加困惑的是，我所做的只是将函数从直接调用变为参数。

下面失败的代码（简化），但是如果我内联runner，即将其作为app的参数删除，则一切正常。

main :: IO ()
main = do
  runSpock 8080 $ app runner

runner :: ReaderT (W.Options -> String -> IO (Response BL.ByteString)) m a -> m a
runner r = runReaderT r W.getWith

app :: (ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a -> IO a) -> IO Middleware
app runner = do
  spockT (runner) $ routes

routes :: MonadIO m => SpockCtxT ctx (ReaderT (W.Options -> String -> IO (Response BL.ByteString)) m) ()
routes = do
  get "healthz" $ text "ok"

Answer 1

我对您使用的库不熟悉，但是spockT的参数必须是多态的。您是说打电话 app的人来确定a是什么。但是spockT希望自己做出决定（可能以多种方式）。具体来说，

spockT :: MonadIO m => (forall a. m a -> IO a) -> SpockT m () -> IO Middleware

尝试将{-# language RankNTypes #-}添加到文件顶部，并将app的签名更改为

app
  :: (forall a. ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a -> IO a)
  -> IO Middleware

更多说明：

spockT的类型如下

spockT :: MonadIO m => (forall a. m a -> IO a) -> ...

这是怎么回事？ MonadIO类看起来像这样

class Monad m => MonadIO m where
  liftIO :: IO a -> m a

这意味着如果您拥有MonadIO m，那么您就拥有一个函数forall a. IO a -> m a。 spockT的函数参数看起来很像，但是却相反。如果m用某种“上下文”包装了IO动作，则该函数必须剥离该上下文，也许还要添加其他信息。

无法将类型a与a1匹配，但类型看起来相同

1 个答案: