【发布时间】：2012-08-30 18:57:28
【问题描述】：
为了熟悉 Haskell 中的 STM，我写了以下解决哲学家就餐问题的方法：
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，似乎不存在明显的并发问题：每个哲学家最终都会吃饭，似乎没有哲学家受到青睐。但是，如果我从philosopher 中删除randomDelay 语句，编译并运行，我的程序的输出如下所示：
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？
【问题讨论】：
-
如果这是作业，请添加作业标签。
-
这不是家庭作业，我在 Real World Haskell 中阅读了有关 STM 的内容，我正在努力熟悉它。
-
使用我的设置（Debian 6 测试，ghc 7.4.1，runhaskell/ghc -O2 --make philosphys.hs）我不认为我有问题 - 哲学家 1 ..8 正在吃东西并轮流思考。你的 ghc 版本是什么，你是在编译还是使用 ghci？
-
我在 Ubuntu 12.04 上使用 ghc 7.4.1。如果我使用
-O2编译，问题仍然存在。
标签： haskell concurrency ghc stm