提前,很抱歉这篇长文。
我正在用 Haskell 编写一个事件驱动的应用程序,因此我需要存储几个回调函数以供进一步使用。我希望这样的回调是:
ReaderT
, ErrorT
, StateT
而不是光秃秃的IO
小号; (MonadIO m, MonadReader MyContext m, MonadState MyState m, MonadError MyError m) => m ()
,而不是 ReaderT MyContext (StateT MyState (ErrorT MyError IO)))
让我们忘记
State
和 Error
层,为了简单起见。我开始编写所有回调的记录,存储在
MyContext
中, 就像是: data MyContext = MyContext { _callbacks :: Callbacks {- etc -} }
-- In this example, 2 callbacks only
data Callbacks = Callbacks {
_callback1 :: IORef (m ()),
_callback2 :: IORef (m ())}
主要问题是:在哪里放置
m
的类型类约束?我尝试了以下,但没有编译:Callbacks
与 m
如 :data (MonadIO m, MonadReader (MyContext m) m) => Callbacks m = Callbacks {
_callback1 :: IORef (m ()),
_callback2 :: IORef (m ())}
如
Callbacks
是 MyContext
的一部分,后者也必须参数化,这会导致无限类型问题(MonadReader (MyContext m) m
)。 data Callbacks = forall m . (MonadIO m, MonadReader MyContext m) => Callbacks {
_callback1 :: IORef (m ()),
_callback2 :: IORef (m ())}
在我编写在
Callbacks
中注册新回调的实际代码之前,它似乎工作正常。 :register :: (MonadIO m, MonadReader MyContext m) => m () -> m ()
register f = do
(Callbacks { _callback1 = ref1 }) <- asks _callbacks -- Note the necessary use of pattern matching
liftIO $ modifyIORef ref1 (const f)
但我收到以下错误(此处简化):
Could not deduce (m ~ m1)
from the context (MonadIO m, MonadReader MyContext m)
bound by the type signature for
register :: (MonadIO m, MonadReader MyContext m) => m () -> m ()
or from (MonadIO m1, MonadReader MyContext m1)
bound by a pattern with constructor
Callbacks :: forall (m :: * -> *).
(MonadIO m, MonadReader MyContext m) =>
IORef (m ())
-> IORef (m ())
-> Callbacks,
Expected type: m1 ()
Actual type: m ()
我找不到解决方法。
如果有人能启发我,我将不胜感激。如果有的话,设计这个的好方法是什么?
预先感谢您的意见。
[编辑] 据我了解 ysdx 的回答,我尝试使用
m
参数化我的数据类型。没有施加任何类型类约束,但后来我无法制作 Callbacks
Data.Default
的一个实例;写这样的东西:instance (MonadIO m, MonadReader (MyContext m) m) => Default (Callbacks m) where
def = Callbacks {
_callback1 = {- something that makes explicit use of the Reader layer -},
_callback2 = return ()}
...导致 GHC 提示:
Variable occurs more often in a constraint than in the instance head
in the constraint: MonadReader (MyContext m) m
它建议使用 UndecidableInstances,但我听说这是一件非常糟糕的事情,虽然我不知道为什么。这是否意味着我必须放弃使用
Data.Default
?
最佳答案
简单的改编(使东西编译):
data MyContext m = MyContext { _callbacks :: Callbacks m }
data Callbacks m = Callbacks {
_callback1 :: IORef (m ()),
_callback2 :: IORef (m ())}
-- Needs FlexibleContexts:
register :: (MonadIO m, MonadReader (MyContext m) m) => m () -> m ()
register f = do
(Callbacks { _callback1 = ref1 }) <- asks _callbacks
liftIO $ modifyIORef ref1 (const f)
但是 -XFlexibleContexts 是必需的。
你真的需要 IORef 吗?为什么不使用简单的状态单子(monad)?
import Control.Monad.State
import Control.Monad.Reader.Class
import Control.Monad.Trans
data Callbacks m = Callbacks {
_callback1 :: m (),
_callback2 :: m ()
}
-- Create a "new" MonadTransformer layer (specialization of StateT):
class Monad m => MonadCallback m where
getCallbacks :: m (Callbacks m)
setCallbacks :: Callbacks m -> m ()
newtype CallbackT m a = CallbackT (StateT (Callbacks (CallbackT m) ) m a)
unwrap (CallbackT x) = x
instance Monad m => Monad (CallbackT m) where
CallbackT x >>= f = CallbackT (x >>= f')
where f' x = unwrap $ f x
return a = CallbackT $ return a
instance Monad m => MonadCallback (CallbackT m) where
getCallbacks = CallbackT $ get
setCallbacks c = CallbackT $ put c
instance MonadIO m => MonadIO (CallbackT m) where
liftIO m = CallbackT $ liftIO m
instance MonadTrans (CallbackT) where
lift m = CallbackT $ lift m
-- TODO, add other instances
-- Helpers:
getCallback1 = do
c <- getCallbacks
return $ _callback1 c
-- This is you "register" function:
setCallback1 :: (Monad m, MonadCallback m) => m () -> m ()
setCallback1 f = do
callbacks <- getCallbacks
setCallbacks $ callbacks { _callback1 = f }
-- Test:
test :: CallbackT IO ()
test = do
c <- getCallbacks
_callback1 c
_callback2 c
main = runCallbackT test s
where s = Callbacks { _callback1 = lift $ print "a" (), _callback2 = lift $ print "b" }
即使没有 MonadIO,此代码代码也可以工作。
定义“默认”似乎工作正常:
instance (MonadIO m, MonadCallback m) => Default (Callbacks m) where
def = Callbacks {
_callback1 = getCallbacks >>= \c -> setCallbacks $ c { _callback2 = _callback1 c },
_callback2 = return ()}
关于haskell - 在 Haskell 中存储多态回调,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12044490/