Re: Help on Understanding Locks in Multithreading

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Re^2: Help on Understanding Locks in Multithreading by koti688 (Sexton) on Feb 25, 2009 at 11:12 UTC
Thanks for the link. But i studied that already and i struck with this example. I am confused how the $foo variable is locked in the functions. Please explain the locking in this code , if you get anything. i am asking very basic things i guess :(	[reply]
Re^3: Help on Understanding Locks in Multithreading by gone2015 (Deacon) on Feb 25, 2009 at 13:04 UTC
What are you expecting, and what did you get which doesn't meet with your expectation ? FWIW, one wrinkle with threads is that the return context for the thread is set at `threads->new` time, not on the context of `$thr->join()`. Your code: `$thr = threads->new(\&sub1, "THREAD1 :"); ... @ReturnData = $thr->join; ...` [download] should be: `($thr) = threads->new(\&sub1, "THREAD1 :"); ... @ReturnData = $thr->join; ...` [download] to provide the right context at the right time. The fragment: `{ lock $foo; $foo=0; }` [download] is reasonably plausible. You're locking `$foo` for just long enough to give it a new value -- so there is no possibility of some other thread attempting to read or change `$foo` while it's in any intermediate state -- provided they too `lock $foo` before doing anything with it. (The lock is dropped at the end of the block it is contained in.) I note that `$foo` is not initialised to anything. I note that you: `print " Final value :$foo\n";` [download] without locking it. But that's probably OK, because `$foo` is only changed by the same thread. I wonder: you're not expecting `$foo` to work as some kind of semaphore, are you ?	[reply] [d/l] [select]
Re^3: Help on Understanding Locks in Multithreading by Wiggins (Hermit) on Feb 25, 2009 at 20:38 UTC
I am confused how the $foo variable is locked in the functions. Actually '$foo' itself isn't locked, execution access is locked. By that I mean that only one thread may be executing in the code blocks that have the 'lock' command at a time. And those blocks just happening to be modifying 'foo' (well, hopefully by design). The locks are controlling/limiting execution flow to a single thread at a time. This stops 2 threads from doing near simultaneous modifications to $foo. In the old days this code was called a 'critical code section'. It is always better to have seen your target for yourself, rather than depend upon someone else's description.	[reply]
Re^3: Help on Understanding Locks in Multithreading by Anonymous Monk on Feb 25, 2009 at 12:20 UTC
Try using smaller numbers, not 1000/5000	[reply]
Re^3: Help on Understanding Locks in Multithreading by Anonymous Monk on Feb 25, 2009 at 12:18 UTC
I am confused how the $foo variable is locked in the functions. Through the magic of perl threads? I'm sorry, I don't understand your question, but I don't think I could explain lock better than perlthrtut. Maybe it would be easier for you to understand if you told us where you got that code, what its supposed to do, etc?	[reply]