我正在努力更深入地了解Haskell中的并发性。我有以下代码:
import Control.Concurrent
main :: IO ()
main = do
arr <- return $ [1..9]
t <- newMVar 1
forkIO (takeMVar t >> (print.show) arr >> putMVar t 1)
forkIO (takeMVar t >> (print.show) arr >> putMVar t 1)
forkIO (takeMVar t >> (print.show) arr >> putMVar t 1)
forkIO (takeMVar t >> (print.show) arr >> putMVar t 1)
forkIO (takeMVar t >> (print.show) arr >> putMVar t 1)
return ()
有时我会看到打印操作重叠,导致出现以下结果(请查看第二个调用):
*Main Control.Concurrent> :l test.hs
[1 of 1] Compiling Main ( test.hs, interpreted )
Ok, modules loaded: Main.
*Main Control.Concurrent> main
"[1,2,3,4,5,6,7,8,9]"
"[1,2,3,4,5,6,7,8,9]"
"[1,2,3,4,5,6,7,8,9]"
"[1,2,3,4,5,6,7,8,9]"
"[1,2,3,4,5,6,7,8,9]"
*Main Control.Concurrent> main
"[1,2,3,4,5,6,7,8,9]"
"[1,2,3,4,5,6,7,8,9]"
"[1,2,3,4,5,6,7,8,9]"
["[1,2,3,4,5,6,7,8,9]"
?"[1,2,3,4,5,6,7,8,9]"
1h*Main Control.Concurrent>
我不明白为什么会发生这种情况。使用MVar处理[1..9]
也不好。
print
=putStrLn.show
。print.show
调用了两次show
,有点冗余。 - jrockway