Question

我想我理解State Monad是如何运作的。我设法写了一些使用State Monad的代码。

我理解Monad实例的运作方式：

instance Monad (State s) where  
return x = State $ \s -> (x,s)  
(State h) >>= f = State $ \s -> let (a, newState) = h s  
                                    (State g) = f a  
                                in  g newState

此代码完美无缺：

type PeopleStack = [String]

enterClub :: String -> State PeopleStack String
enterClub name = state $ \xs -> (name ++ " entered club", name:xs)

leaveClub :: State PeopleStack String
leaveClub = state $ \(x:xs) -> ("Someone left the club", xs)

clubAction :: State PeopleStack String
clubAction = do
  enterClub "Jose"
  enterClub "Thais"
  leaveClub
  enterClub "Manuel"

然而，当我尝试在绑定函数中编写clubAction时，我似乎无法使它工作。

这就是我的尝试：

let statefulComputation1 = enterClub "Jose"
statefulComputation1 :: State PeopleStack String

runState (statefulComputation1 >>= (enterClub "Manuel") >>= leaveClub) []

我收到此错误：

<interactive>:13:22:
Couldn't match type ‘StateT
                       PeopleStack Data.Functor.Identity.Identity String’
               with ‘String
                     -> StateT PeopleStack Data.Functor.Identity.Identity a’
Expected type: String
               -> StateT PeopleStack Data.Functor.Identity.Identity a
  Actual type: State PeopleStack String
Relevant bindings include
  it :: (a, PeopleStack) (bound at <interactive>:13:1)
In the second argument of ‘(>>=)’, namely ‘leaveClub’
In the first argument of ‘runState’, namely
  ‘(state1 >>= leaveClub)’

我的问题是如何使用bind将该表示法转换为函数。

Answer 1

您需要使用(>>)代替(>>=)：

runState (statefulComputation1 >> (enterClub "Manuel") >> leaveClub) []

(enterClub "Manuel")的类型为State PeopleStack String，而(>>=)则需要函数String -> State PeopleStack a作为其第二个参数。由于您不能使用statefulComputation1的结果，因此可以将它们与(>>)结合使用，忽略第一个状态计算的结果。

Answer 2

绑定运算符右侧的每个项目（＆gt;＆gt; =）都需要是一个lambda ....，这样就可以了

runState (statefulComputation1 >>= \_ -> enterClub "Manuel" >>= \_ -> leaveClub) []

或者，您可以使用简写(>>)

runState (statefulComputation1 >> enterClub "Manuel" >> leaveClub) []

State Monad Bind

2 个答案: