如何在两个 monad 之间共享状态？
我遇到了以下 monad 问题：
假设我有一个标准 monad 状态，状态为 S = (LS, RS)。我还有另一个 monad：
newtype StateP a = StateP {runP :: S -> (a, RS)}
我想使用 StateP 执行一些计算，然后将状态与 State monad 中的状态合并：
merge m :: StateP() -> State()
merge m = do
s@(l,r) <- get
put (l, snd (runP m s))
它不起作用，但我不明白为什么？还有其他方法可以实现这样的功能吗？
I am stuck with following monad problem:
Let's say I have a standard monad State with state S = (LS, RS). I also have another monad:
newtype StateP a = StateP {runP :: S -> (a, RS)}
I want to perform some computation using StateP then merge state with state in State monad:
merge m :: StateP() -> State()
merge m = do
s@(l,r) <- get
put (l, snd (runP m s))
It is not working, but I don't get why? Is there another way to achieve such functionality?
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如果您的代码几乎正确，请尝试
但您确实需要向我们提供更多详细信息。
On the off-chance that your code is nearly right, try
But you need to give us more details really.
您可以使用 monad 转换器，使用两个 monad 堆栈更明确地对这些需求进行建模：一个只能读取
LS，另一个可以读取和写入LS。要在
ReadWriteLS中运行ReadOnlyLS，我们只需从最外层的状态层中提取出LS，并将其交给内部计算的读取层即可并将计算结果提升回外部 monad：You can use monad transformers to model these requirements more explicitly using two monad stacks: one that can only read
LSand one that can both read and writeLS.To run a
ReadOnlyLSwithinReadWriteLS, we just need to extractLSfrom the outermost state layer, give it to the reader layer of the inner computation and lift the resulting computation back into the outer monad:RunP的功能是如何实现的？你是否为它重新定义了一个 monad 实例并拥有 getP/putP ？你的代码看起来不错，你能提供你使用的m吗？你有什么样的不当行为？
How do you achieve the functionality of RunP? Do you redefine a monad instance for it and have getP/putP? Your code seems fine, can you provide the m you use? What kind of misbehaviour do you get?