我刚刚重新发明了一些monad,但我不确定哪一个。它可以让你对一个计算的步骤进行建模,这样你就可以交错计算大量的步骤,找出哪一步先完成。Haskell:我刚刚改造了哪个monad?
{-# LANGUAGE ExistentialQuantification #-}
module Computation where
-- model the steps of a computation
data Computation a = forall b. Step b (b -> Computation a) | Done a
instance Monad Computation where
(Step b g) >>= f = Step b $ (>>=f) . g
(Done b) >>= f = Step b f
return = Done
runComputation :: Computation a -> a
runComputation (Step b g) = runComputation (g b)
runComputation (Done a) = a
isDone :: Computation a -> Bool
isDone (Done _) = True
isDone _ = False
-- an order for a set of computations
data Schedule a = a :> Computation (Schedule a) | Last
toList :: Schedule a -> [a]
toList Last = []
toList (a :> c) = a : (toList . runComputation) c
-- given a set of computations, find a schedule to generate all their results
type Strategy a = [Computation a] -> Computation (Schedule a)
-- schedule all the completed computations, and step the rest,
-- passing the remaining to the given function
scheduleOrStep :: (Queue (Computation a) -> Computation (Schedule a)) -> Strategy a
scheduleOrStep s cs = scheduleOrStep' id cs
where scheduleOrStep' q ((Done a):cs) = Done $ a :> scheduleOrStep' q cs
scheduleOrStep' q ((Step b g):cs) = scheduleOrStep' (q . (g b:)) cs
scheduleOrStep' q [] = s q
-- schedule all completed compuations, step all the rest once, and repeat
-- (may never complete for infinite lists)
-- checking each row of
-- [ [ c0s0, c1s0, c2s0, ... ]
-- , [ c0s1, c1s1, c2s1, ... ]
-- , [ c0s2, c1s2, c2s2, ... ]
-- ...
-- ]
-- (where cNsM is computation N stepped M times)
fair :: Strategy a
fair [] = Done Last
fair cs = scheduleOrStep (fair . ($[])) cs
-- schedule more steps for earlier computations rather than later computations
-- (works on infinite lists)
-- checking the sw-ne diagonals of
-- [ [ c0s0, c1s0, c2s0, ... ]
-- , [ c0s1, c1s1, c2s1, ... ]
-- , [ c0s2, c1s2, c2s2, ... ]
-- ...
-- ]
-- (where cNsM is computation N stepped M times)
diag :: Enqueue (Computation a)-> Strategy a
diag _ [] = Done Last
diag enq cs = diag' cs id
where diag' (c:cs) q = scheduleOrStep (diag' cs) (enq c q $ [])
diag' [] q = fair (q [])
-- diagonal downwards :
-- [ c0s0,
-- c1s0, c0s1,
-- c2s0, c1s1, c0s2,
-- ...
-- cNs0, c{N-1}s1, ..., c1s{N-1}, c0sN,
-- ...
-- ]
diagd :: Strategy a
diagd = diag prepend
-- diagonal upwards :
-- [ c0s0,
-- c0s1, c1s0,
-- c0s2, c1s1, c2s0,
-- ...
-- c0sN, c1s{N-1}, ..., c{s1N-1}, cNs0,
-- ...
-- ]
diagu :: Strategy a
diagu = diag append
-- a queue type
type Queue a = [a] -> [a]
type Enqueue a = a -> Queue a -> Queue a
append :: Enqueue a
append x q = q . (x:)
prepend :: Enqueue a
prepend x q = (x:) . q
我觉得这可能是某种线程monad?
哈斯克尔是我所知道的唯一一种语言,你无法分辨出你刚刚重新创建了哪个轮子... –
我即将关闭_too localized_,但人们是否真的花时间知道他们正在重塑Haskell中的东西但不是他们正在重塑的问题,使这个问题变得合法(假设很多人最终会重塑这个确切的东西,不管它是什么)? – Mat
@Mat:是的,其实。至少在某些方面。在Haskell中,人们偶尔会做出一些不太有趣的事情,只要给出足够的泛型代码,如果它的类型检查几乎肯定会做一些有用的事情,即使你不确定是什么。这有点类似,因为如果你发明了某些东西来解决某个具体问题,并且它很容易理解,那么很可能已经完成了。当我第一次学习Haskell时,至少有一次我推广了一个特定的解决方案,只是意识到我已经改造了标准库中一个不起眼的角落。 –