Question

为了熟悉Haskell中的STM，我为Dining Philosophers问题写了以下解决方案：

import Control.Concurrent
import Control.Concurrent.STM
import Control.Monad
import System.Random

type Fork = TVar Bool
type StringBuffer = TChan String

philosopherNames :: [String]
philosopherNames = map show ([1..] :: [Int])

logThinking :: String -> StringBuffer -> STM ()
logThinking name buffer = writeTChan buffer $ name ++ " is thinking..."

logEating :: String -> StringBuffer -> STM ()
logEating name buffer = writeTChan buffer $ name ++ " is eating..."

firstLogEntry :: StringBuffer -> STM String
firstLogEntry buffer = do empty <- isEmptyTChan buffer
                          if empty then retry
                                   else readTChan buffer

takeForks :: Fork -> Fork -> STM ()
takeForks left right = do leftUsed <- readTVar left
                          rightUsed <- readTVar right
                          if leftUsed || rightUsed
                             then retry
                             else do writeTVar left True
                                     writeTVar right True

putForks :: Fork -> Fork -> STM ()
putForks left right = do writeTVar left False
                         writeTVar right False

philosopher :: String -> StringBuffer -> Fork -> Fork -> IO ()
philosopher name out left right = do atomically $ logThinking name out
                                     randomDelay
                                     atomically $ takeForks left right
                                     atomically $ logEating name out
                                     randomDelay
                                     atomically $ putForks left right

randomDelay :: IO ()
randomDelay = do delay <- getStdRandom(randomR (1,3))
                 threadDelay (delay * 1000000)

main :: IO ()
main = do let n = 8
          forks <- replicateM n $ newTVarIO False
          buffer <- newTChanIO
          forM_ [0 .. n - 1] $ \i ->
              do let left = forks !! i
                     right = forks !! ((i + 1) `mod` n)
                     name = philosopherNames !! i
                 forkIO $ forever $ philosopher name buffer left right

          forever $ do str <- atomically $ firstLogEntry buffer
                       putStrLn str

当我编译并运行我的解决方案时，似乎没有明显的并发问题：每个哲学家最终都会吃掉，而哲学家似乎并没有受到青睐。但是，如果我从randomDelay删除philosopher语句，编译并运行，我的程序输出如下所示：

1 is thinking...
1 is eating...
1 is thinking...
1 is eating...
2 is thinking...
2 is eating...
2 is thinking...
2 is eating...
2 is thinking...
2 is eating...
2 is thinking...

About 2500 lines later...

2 is thinking...
2 is eating...
2 is thinking...
3 is thinking...
3 is eating...
3 is thinking...
3 is eating...

And so on...

在这种情况下发生了什么？

Answer 1

您需要使用线程运行时编译它并启用rtsopts，并使用+RTS -N（或+RTS -Nk运行它，其中k是线程数。，我得到像

的输出

8 is eating...
6 is eating...
4 is thinking...
6 is thinking...
4 is eating...
7 is eating...
8 is thinking...
4 is thinking...
7 is thinking...
8 is eating...
4 is eating...
4 is thinking...
4 is eating...
6 is eating...
4 is thinking...

关键是，对于另一位思考/吃饭的哲学家来说，如果您的处置中没有多个硬件线程，则必须进行上下文切换。这种上下文切换在这里并不经常发生，在这里没有进行太多的分配，因此每个哲学家都有很多时间在下一轮出现之前思考和吃很多东西。

在你的处置中有足够的线索，所有哲学家都可以同时尝试伸手去拿叉子。

“饮食哲学家”的以下解决方案有什么问题？

1 个答案: