[Haskell-cafe] Concurrency question

Dean Herington heringtonlacey at mindspring.com
Mon Sep 5 02:49:28 EDT 2005


At 9:45 PM +0400 9/4/05, Dmitry Vyal wrote:
>Donald Bruce Stewart wrote:
>
>>Maybe your loop does no allocations, so the scheduler can't get in and do a
>>context switch.  You could put the computation in an external 
>>program, and run
>>it over a fork, using unix signals in the external program to kill the
>>computation after a period of time.
>
>I thought about doing that, but function is closely connected with 
>the rest of the program. Running it in another process would require 
>some parsing of its arguments and I want circumvent these 
>difficulties.
>
>Moreover, this function indeed allocates plenty of memory (creates 
>long lists), so It's just curiously for me to establish the reason 
>of this (mis)behavior. By the way, what does it mean precisely, "no 
>allocations".
>
>This is the top part of program I have trouble with. "resolve" is 
>that sluggish function, which execution I'm trying to break. It hogs 
>a lot of memory, so context switching should occur regular.
>
>I'm new to Haskell, so probably I've just made some really stupid mistake.
>
>Thanks a lot for your help.

I believe you're just observing lazy evaluation at work.  The IO 
computation that you're forking is (return $ resolve cnf).  `resolve` 
is a pure function.  Hence the forked computation succeeds 
immediately--and the thread terminates (successfully)--without 
evaluating (resolve cnf).  It isn't until the case arm that begins 
"Just (ans, stats) ->" that the result of (resolve cnf) is demanded 
and hence evaluation of (resolve cnf) begins.  But this is too late 
for the timeout to have the intended effect.

How to fix?  You need to demand (enough of) the result of (resolve 
cnf) before returning from the IO computation.  What "enough of" 
means depends on how `resolve` is written.  You may find the DeepSeq 
module I wrote (see 
http://www.mail-archive.com/haskell@haskell.org/msg15819.html) 
helpful.

Dean

>
>
>res_timeout=1000000 -- time quota in microseconds
>
>forever a = a >> forever a
>
>main :: IO ()
>main = do args <- getArgs
>	  if (length args /= 1) then usage
>	     else do axioms <- readFile (head args)
>		     let tree = parseInput axioms
>                      case tree of
>                           (Right exprs) ->
>                               do let cnf = normalize $
>					   concatMap to_cnf exprs
>                                  forever $ one_cycle cnf
>			  (Left er) -> putStr $ show er
>
>usage = putStr "usage: resolution <filename>\n"
>
>one_cycle :: CNF -> IO ()
>one_cycle base =
>     do inp <- getLine
>        let lex_tree = parseInput inp
>        case lex_tree of
>            (Right exprs) -> run_resolution $
>                             normalize $ to_cnf (Not (head exprs))
>					++ base
>            (Left er) -> putStr $ show er
>
>-- Here I start a heavy computation
>
>run_resolution :: CNF -> IO ()
>run_resolution cnf =
>     do res <- timeout res_timeout (return $ resolve cnf)
>        case res of
>             Just (ans, stats) -> do print stats
>				    print ans
>             Nothing -> print "***timeout***"
>
>-- These useful subroutines I saw in "Tackling The Awkward Squad"
>
>par_io :: IO a -> IO a -> IO a
>par_io t1 t2 = do c <- newEmptyMVar :: IO (MVar a)
>                   id1 <- forkIO $ wrapper c t1
>                   id2 <- forkIO $ wrapper c t2
>                   res <- takeMVar c
>                   killThread id1
>                   killThread id2
>                   return res
>     where wrapper :: MVar a -> IO a -> IO ()
>           wrapper mvar io = do res <- io
>                                putMVar mvar res
>
>timeout :: Int -> IO a -> IO (Maybe a)
>timeout n t = do res <- par_io timer thr
>                  return res
>     where thr = do res <- t
>                    return $ Just res
>           timer = do threadDelay n
>                      return Nothing
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