我为语句创建了简单的计算器。我想使用变压器来实现这一点-将IO monad与State混合
有人能解释一下怎么做吗?这是我无法处理的事情——变形金刚。
execStmt :: Stmt -> State (Map.Map String Int) ()
execStmt s = case s of
SAssigment v e -> get >>= (s -> put (Map.insert v (fromJust (runReader (evalExpM e) s)) s))
SIf e s1 s2 -> get >>= (s -> case (fromJust (runReader (evalExpM e) s)) of
0 -> execStmt s2
_ -> execStmt s1
)
SSkip -> return ()
SSequence s1 s2 -> get >>= (s -> (execStmt s1) >>= (s' -> execStmt s2))
SWhile e s1 -> get >>= (s -> case (fromJust (runReader (evalExpM e) s)) of
0 -> return ()
_ -> (execStmt s1) >>= (s' -> execStmt (SWhile e s1)))
execStmt' :: Stmt -> IO ()
execStmt' stmt = putStrLn $ show $ snd $ runState (execStmt stmt) Map.empty
以下是的基本程序大纲
newtype StateIO s a = SIO {runSIO :: s -> IO (a, s)}
put :: s -> StateIO s ()
put s' = SIO $ _s -> return ((), s')
liftIO :: IO a -> StateIO s a
liftIO ia = SIO $ s -> do
a <- ia
return (a, s)
instance Functor (StateIO s) where
fmap ab (SIO sa) = SIO $ s -> do
(a, s') <- sa s
let b = ab a
return (b, s')
instance Applicative (StateIO s) where
pure a = SIO $ s -> return (a, s)
(SIO sab) <*> (SIO sa) = SIO $ s -> do
(ab, s' ) <- sab s
(a , s'') <- sa s'
let b = ab a
return (b, s')
StateIO s a
是接受输入状态(类型为s
)的东西,并返回IO操作以产生类型为a
的东西以及新状态。
要检查是否理解,请执行以下
- 写入一个值
get :: StateIO s s
,用于检索状态 - 为
Monad (StateIO s)
编写一个实例(它将类似于上面的代码) - 最后,这是理解transformers的重要一步,就是定义
newtype StateT m s a = StateT {run :: s -> m (a, s)}
,并将上面的代码转换为Monad m
。这将向您展示monad转换器是如何工作的