Haskell-无法使用类似Monad的定义来定义类似于State Monad的函数 [英] Haskell - Unable to define a State monad like function using a Monad like definition
问题描述
我试图通过尝试编写泛型版本的功能来理解Monad的概念,这些功能随后可能包含记录日志,更改状态的副作用.
I am trying to understand the concept of Monad by attempting to write generic version of functions that might be then include side effects to log, change state.
这是我想出的:(代码有点长,但是它可以显示我是如何理解monad的-这种方法可能不正确)
Here is what I came up with: (The code is bit long, but it is there to show how I approached understanding monad - and this approach may not be correct)
data Maybe' a = Nothing' | Just' a deriving Show
sqrt' :: (Floating a, Ord a) => a -> Maybe' a
sqrt' x = if x < 0 then Nothing' else Just' (sqrt x)
inv' :: (Floating a, Ord a) => a -> Maybe' a
inv' x = if x == 0 then Nothing' else Just' (1/x)
log' :: (Floating a, Ord a) => a -> Maybe' a
log' x = if x == 0 then Nothing' else Just' (log x)
sqrtInvLog' :: (Floating a, Ord a) => a -> Maybe' a
sqrtInvLog' x = case (sqrt' x) of
Nothing' -> Nothing'
(Just' y) -> case (inv' y) of
Nothing' -> Nothing'
(Just' z) -> log' z
-- Now attempt to simplify the nested case:
fMaybe' :: (Maybe' a) -> (a -> Maybe' b) -> Maybe' b
fMaybe' Nothing' _ = Nothing'
fMaybe' (Just' x) f = f x
-- using fMaybe':
sqrtInvLog'' :: (Floating a, Ord a) => a -> Maybe' a
sqrtInvLog'' x = (sqrt' x) `fMaybe'` (inv') `fMaybe'` (log')
-- now we can generalize the concept to any type, instead of just Maybe' by defining a Monad =>
class Monad' m where
bind' :: m a -> (a -> m b) -> m b
return' :: a -> m a
instance Monad' Maybe' where
bind' Nothing' _ = Nothing'
bind' (Just' x) f = f x
return' x = Just' x
-- using Monad sqrtInvLog'' can be written as:
sqrtInvLog''' :: (Floating a, Ord a) => a -> Maybe' a
sqrtInvLog''' x = (sqrt' x) `bind'` (inv') `bind'` (log')
-- Further lets attempt to use this for state maintenence and logging, logging:
-- first attempt the specific version:
data ST a = ST (a, Maybe' a) deriving Show
sqrtSt :: (Floating a, Ord a)=> a -> a -> ST a
sqrtSt st x = let r = sqrt' x in case r of
Nothing' -> ST (st, Nothing')
(Just' y) -> ST (st+y, (Just' y))
invSt :: (Floating a, Ord a)=> a -> a -> ST a
invSt st x = let r = inv' x in case r of
Nothing' -> ST (st, Nothing')
(Just' y) -> ST (st+y, (Just' y))
logSt :: (Floating a, Ord a)=> a -> a -> ST a
logSt st x = let r = log' x in case r of
Nothing' -> ST (st, Nothing')
(Just' y) -> ST (st+y, (Just' y))
-- let us first define function which is similar to bind and manipulates the state and invokes the given function:
stBind :: (Floating a, Ord a) => ST a -> (a->a->ST a) -> ST a
stBind (ST (a, Nothing')) _ = ST (a, Nothing')
stBind (ST (s, (Just' y))) f = f s y
sqrtInvLogSt :: (Floating a, Ord a) => a -> a -> ST a
sqrtInvLogSt st x = (sqrtSt st x) `stBind` (invSt) `stBind` (logSt)
-- stBind does not fit the pattern of bind
-- Another version:
sqrtSt' :: (Floating a, Ord a)=> ST a -> ST a
sqrtSt' (ST (st, Nothing')) = ST (st, Nothing')
sqrtSt' (ST (st, (Just' x))) = let r = sqrt' x in case r of
Nothing' -> ST (st, Nothing')
(Just' y) -> ST (st+y, (Just' y))
invSt' :: (Floating a, Ord a)=> ST a -> ST a
invSt' (ST (st, Nothing')) = ST (st, Nothing')
invSt' (ST (st, (Just' x))) = let r = inv' x in case r of
Nothing' -> ST (st, Nothing')
(Just' y) -> ST (st+y, (Just' y))
logSt' :: (Floating a, Ord a)=> ST a -> ST a
logSt' (ST (st, Nothing')) = ST (st, Nothing')
logSt' (ST (st, (Just' x))) = let r = log' x in case r of
Nothing' -> ST (st, Nothing')
(Just' y) -> ST (st+y, (Just' y))
-- define stBind' here
stBind' :: (Floating a, Ord a) => ST a -> (ST a->ST a) -> ST a
stBind' (ST (a, Nothing')) _ = ST (a, Nothing')
stBind' stx f = f stx
sqrtInvLogSt' :: (Floating a, Ord a) => ST a->ST a
sqrtInvLogSt' stx = (sqrtSt' stx) `stBind'` (invSt') `stBind'` (logSt')
-- Even this does not fit the pattern of bind,
最后定义的函数stBind'不适合bind'模式. 在这种情况下如何提出实现以匹配绑定运算符签名?
The function stBind' defined in the end does not fit the pattern of bind'. How can I come up with an implementation in this situation to match the bind operator signature?
推荐答案
冒着被破坏的风险,使State
monad的定义达到顶峰可能很有趣,因为它在所有东西都变成变压器之前就已经存在:>
At the risk of spoilers, it might be interesting to peak at the definition of the State
monad as it existed before everything became transformer-ified:
newtype State s a = State { runState :: s -> (a, s) }
也就是说,状态为s
的有状态动作会产生类型为a
的值,它是从旧状态s
到值新状态(a, s)
的函数.
That is: a stateful action with state s
that produces a value of type a
is a function from the old state s
to a value and a new state, (a, s)
.
从正确的State
定义开始,应该使开发的其余部分更容易迷惑.
Starting from the correct definition for State
should make the remainder of your development easier to puzzle through.
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