Question

我正在玩国家队和队列。目前我有以下代码：

{-# LANGUAGE ViewPatterns, FlexibleContexts #-}
module Main where

import Criterion.Main
import Control.Monad.State.Lazy
import Data.Maybe (fromJust)
import Data.Sequence ((<|), ViewR ((:>)))
import qualified Data.Sequence as S

--------------------------------------------------------
data Queue a = Queue { enqueue :: [a], dequeue :: [a] }
                                        deriving (Eq, Show)
-- adds an item
push :: a -> Queue a -> Queue a
push a q = Queue (a:enqueue q) (dequeue q)

pop :: Queue a -> Maybe (a, Queue a)
pop q = if null (dequeue q) then
          go $ Queue [] (reverse (enqueue q))
        else
          go q
  where go (Queue _ []) = Nothing
        go (Queue en (x:de)) = Just (x, Queue en de)

queueTst :: Int -> Queue Int -> Queue Int
queueTst 0 q = q
queueTst n q | even n =  queueTst (n - 1) (push (100 + n) q)
             | otherwise = queueTst (n - 1)
                           (if popped == Nothing then q
                            else snd (fromJust popped))
    where popped = pop q
-------------------------------------------------------------
pushS :: a -> S.Seq a -> S.Seq a
pushS a s = a <| s

pushS' :: a -> State (S.Seq a) (Maybe a)
pushS' a = do
  s <- get
  put (a <| s)
  return Nothing

pushS'' :: a -> State (S.Seq a) (Maybe a)
pushS'' a = get >>= (\g -> put (a <| g)) >> return Nothing

popS :: S.Seq a -> Maybe (a, S.Seq a)
popS (S.viewr -> S.EmptyR) = Nothing
popS (S.viewr -> s:>r) = Just (r,s)

popS' :: State (S.Seq a) (Maybe a)
popS' = do
  se <- get
  let sl = popS'' se
  put $ snd sl
  return $ fst sl
  where popS'' (S.viewr -> S.EmptyR) = (Nothing, S.empty)
        popS'' (S.viewr -> beg:>r) = (Just r, beg)

queueTstS :: Int -> S.Seq Int -> S.Seq Int
queueTstS 0 s = s
queueTstS n s | even n = queueTstS (n - 1) (pushS (100 + n) s)
              | otherwise = queueTstS (n - 1)
                            (if popped == Nothing then s
                             else snd (fromJust popped))
      where popped = popS s

queueTstST :: Int -> State (S.Seq Int) (Maybe Int)
queueTstST n =
  if (n > 0) then
     if even n then
       pushS' (100 + n) >> queueTstST (n - 1)
     else
       popS' >> queueTstST (n - 1)
  else return Nothing

main :: IO ()
main = defaultMain
  [ bench "Twin Queue" $ whnf (queueTst 550) (Queue [500,499..1] [])
  , bench "Sequence Queue" $ whnf (queueTstS 550) (S.fromList [500,499..1])
  , bench "State Queue" $ whnf
                  (runState (queueTstST 550)) (S.fromList [500,499..1])
  ]

这是一些代码，但实际上这里唯一相关的功能是main和queueTstST。有没有办法退出queueTstST，同时保留最后一个“可能值”而不是“没有”？

Answer 1

queueTstST :: Int -> State (S.Seq Int) (Maybe Int)
queueTstST n =
  if (n > 1) then
     if even n then
       pushS' (100 + n) >> queueTstST (n - 1)
     else
       popS' >> queueTstST (n - 1)
  else popS'

Answer 2

如果在递归函数中添加参数，则可以记住最后一个值。

np.exp(-(t-5)/tauN1))

从State Monad Loop出口

2 个答案: