Haskell 中状态的函子/应用实例
在阅读（并浏览了 Wadler 关于 monad 的论文的某些部分）后，我决定更仔细地研究这篇论文，为他描述的每个 monad 定义函子和应用实例。 类型同义词
type M a = State -> (a, State)
type State = Int
使用Wadler 用来定义状态 monad 的
fmap' :: (a -> b) -> M a -> M b
fmap' f m = \st -> let (a, s) = m st in (f a, s)
pure' :: a -> M a
pure' a = \st -> (a, st)
(<@>) :: M (a -> b) -> M a -> M b
sf <@> sv = \st -> let (f, st1) = sf st
(a, st2) = sv st1
in (f a, st2)
return' :: a -> M a
return' a = pure' a
bind :: M a -> (a -> M b) -> M b
m `bind` f = \st -> let (a, st1) = m st
(b, st2) = f a st1
in (b, st2)
，我有以下内容（使用相关名称，以便稍后可以使用 newtype 声明来定义它们）。例如，当我切换到在 newtype 声明中使用类型构造函数时，
newtype S a = S (State -> (a, State))
一切都会崩溃。例如，一切都只是轻微的修改，
instance Functor S where
fmap f (S m) = S (\st -> let (a, s) = m st in (f a, s))
instance Applicative S where
pure a = S (\st -> (a, st))
但是由于 lambda 表达式隐藏在该类型构造函数中，GHC 中什么也没有运行。现在我看到的唯一解决方案是定义一个函数：
isntThisAnnoying s (S m) = m s
为了将 s 绑定到“st”并实际返回一个值，例如，
fmap f m = S (\st -> let (a, s) = isntThisAnnoying st m in (f a, s))
是否有另一种不使用这些辅助函数的方法来做到这一点？
After reading (and skimming some sections of) Wadler's paper on monads, I decided to work through the paper more closely, defining functor and applicative instances for each of the monads he describes. Using the type synonym
type M a = State -> (a, State)
type State = Int
Wadler uses to define the state monad, I have the following (using related names so I can define them with a newtype declaration later on).
fmap' :: (a -> b) -> M a -> M b
fmap' f m = \st -> let (a, s) = m st in (f a, s)
pure' :: a -> M a
pure' a = \st -> (a, st)
(<@>) :: M (a -> b) -> M a -> M b
sf <@> sv = \st -> let (f, st1) = sf st
(a, st2) = sv st1
in (f a, st2)
return' :: a -> M a
return' a = pure' a
bind :: M a -> (a -> M b) -> M b
m `bind` f = \st -> let (a, st1) = m st
(b, st2) = f a st1
in (b, st2)
When I switch to using a type constructor in a newtype declaration, e.g.,
newtype S a = S (State -> (a, State))
everything falls apart. Everything is just a slight modification, for instance,
instance Functor S where
fmap f (S m) = S (\st -> let (a, s) = m st in (f a, s))
instance Applicative S where
pure a = S (\st -> (a, st))
however nothing runs in GHC due to the fact that the lambda expression is hidden inside that type constructor. Now the only solution I see is to define a function:
isntThisAnnoying s (S m) = m s
in order to bind s to 'st' and actually return a value, e.g.,
fmap f m = S (\st -> let (a, s) = isntThisAnnoying st m in (f a, s))
Is there another way to do this that doesn't use these auxiliary functions?
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如果您查看此处，您会看到他们是这样定义的：
以便给内部 lambda 命名。
If you look here, you will see that they define it this way:
so as to give the inner lambda a name.
通常的方法是定义 newtype newtype S a = S {runState : State -> （a，状态）}。然后，您可以编写
runState t s而不是isntThisAnnoying s (S m)，其中t与S m相同>.您必须使用
newtype因为类型同义词不能是类型类实例。The usual way is to define
newtype newtype S a = S {runState : State -> (a, State)}. Then instead of yourisntThisAnnoying s (S m)you can writerunState t swheretis the same asS m.You have to use a
newtypebecause type synonyms cannot be typeclass instances.