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无法将类型A与A1匹配,但类型看起来相同

haskell

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Jesse Shieh José Valim · 技术社区 · 6 年前

我在学习haskell,编译器给了我一个我不太明白的错误。它说预期类型和实际类型不匹配,但对我来说,它们看起来是相同的。有人能帮我理解这个错误想说什么吗?

/app/app/Main.hs:75:11: error:
    â¢ Couldn't match type âaâ with âa1â
      âaâ 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
      âa1â 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
    â¢ In the first argument of âspockTâ, namely â(r)â
      In the expression: spockT (r)
      In a stmt of a 'do' block: spockT (r) $ routes apiKey template
    â¢ 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
   |           ^

更让我困惑的是,我所做的只是将一个函数从被直接调用变成一个参数。

失败的代码在下面(简化),但是如果我内联 runner 也就是说,将其作为 app 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"

1 回复 | 直到 6 年前

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4

dfeuer 6 年前

我对你正在使用的图书馆不熟悉,但是 spockT 必须是多态的。你是说那个人电话 app 决定什么 a 是。但是 斯科克特 想要自己做出决定(可能有多种方式)。明确地,

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

尝试添加 {-# language RankNTypes #-} 到文件的顶部并更改 应用程序 到

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

再解释一下:

类型 斯科克特 看起来像

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

这是怎么回事?这个 MonadIO 类看起来像这样

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

这意味着如果你有 MonadIO m ,那么你就有了一个函数 forall a. IO a -> m a . 函数参数 斯科克特 看起来很像,但情况恰恰相反。如果 m 包装一个 IO 操作时使用某种“上下文”,那么函数必须除去该上下文,可能还需要添加其他信息。