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的国家实例如何运作:
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
这段代码非常好用:
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"
但是当我尝试在绑定函数中编写clubAction时,我似乎无法使其工作 .
这是我试过的:
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)’
我的问题是我如何使用bind将该表示法转换为函数 .
2 回答
绑定运算符(>> =)右侧的每个项都需要是一个lambda ....,这样就行了
或者,你可以用速记
(>>)
您需要使用
(>>)
而不是(>>=)
:(enterClub "Manuel")
的类型为State PeopleStack String
,而(>>=)
需要函数String -> State PeopleStack a
作为其第二个参数 . 由于您不使用statefulComputation1
的结果,您可以将它们与(>>)
组合,忽略第一个状态计算的结果 .