Haskell 改变 IO 函数的状态

Question

我正在尝试更新 IO 函数中的记录。我尝试使用状态单子(monad)但没有成功。我在某处看到一条评论说可以用状态单子(monad)解决类似的问题。是否可以使用 State-monad 来执行此操作，或者我是否必须使用 monad 转换器？如果是这样，我将不胜感激有关如何做到这一点的简短解释。

{-# LANGUAGE GADTs #-}

data CarState = CS {color :: String  }

initState :: CarState
initState = CS {color = "white"}


data Car where
   ChangeColor :: String  -> Car

func :: Car -> CarState -> IO ()
func (ChangeColor col) cs = do

     putStrLn ("car is " ++ show(color cs))

     -- Change state here somehow
    --  colorChange "green"

     putStrLn ("car is now" ++ show(color cs))


main :: IO ()
main = func (ChangeColor "green") initState 

colorChange :: String -> MyState CarState String
colorChange col = do
     cs <- get
     put $ cs {color = col}
     return col


data MyState s a = St (s -> (a,s)) 

runSt :: MyState s a -> s -> (a,s)
runSt (St f) s = f s 

evalSt :: MyState s a -> s -> a
evalSt act = fst . runSt act 

get :: MyState s s
get = St $ \s -> (s,s)

put :: s -> MyState s ()
put s = St $ \_ -> ((),s)

instance Monad (MyState s) where
   return x       = St $ \s -> (x,s) 
   (St m) >>= k = St $ \s_1 -> let (a, s_2) = m s_1
                               St k_m = k a 
                           in k_m s_2 

instance Functor (MyState s) where
    fmap f (St g) = St $ \s0 -> let (a, s1) = g s0
                                in (f a, s1)

instance Applicative (MyState s) where
    pure a = St (\s -> (a,s))
    (<*>) (St sa) (St sb) = St (\s0 -> let (fn, s1) = sa s0 
                                       (a,  s2) = sb s1  
                                   in (fn a, s2))

Answer 1

在 Haskell 中，你永远无法通过任何机制就地改变数据。完全没有。¹

Haskell 中状态更新建模的方式是生成一个新值，该值与旧值相同，除了“更新”位之外。

例如，您想要做的事情可以这样表达:

func :: Car -> CarState -> CarState
func (ChangeColor c) cs = cs { color = c }

<小时/>

¹ 好吧，技术上有时可以，但它只适用于预先安排为可变的数据，而且通常相当麻烦。这不是编写程序的正常方式。