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[Caml-list] Resource acquisition is initialization
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| Date: | -- (:) |
| From: | Brian Hurt <brian.hurt@q...> |
| Subject: | Re: [Caml-list] Resource acquisition is initialization |
On Thu, 12 Dec 2002, Alessandro Baretta wrote:
>
>
> Brian Hurt wrote:
> > On Wed, 11 Dec 2002, Blair Zajac wrote:
>
> > The only thing I can think of which cannot tolerate the cleanup slop is
> > releasing a mutex. Deleting the mutex itself is a good application for
> > GC, unlocking the mutex is not.
>
> It is not the GC which unlocks the mutex but the application
> code. See Mutex.unlock .
Calling a function to unlock a mutex is precise- so long as you remember
to always unlock the mutex. At least, as I'm understanding things. It's
the difference between (in Java-esque):
{
mutex.lock();
/* do stuff needing the mutex to be locked */
mutex.unlock();
}
and:
{
mutex_lock lock_obj = mutex.get_lock();
/* do stuff needing the mutex to be locked */
}
/* lock_obj is now garbage- eventually the GC will clean it up, thus
* unlocking the mutex. But this may be a while. Until that happens,
* everyone still thinks we're holding the mutex locked.
*/
The problem with the first representation is that you can forget to unlock
the mutex, especially on error paths. Consider:
{
mutex.lock();
/* do stuff */
if (error) { throw AnException(); }
/* Opps- did I just forget to unlock the mutex? */
mutex.unlock();
}
The problem with the second way is that you continue holding the lock for
some indefinate amount of time until the GC frees the object. This is
what I meant by "the time you hold the lock is defined by the lifetime of
an object".
Note that for most other resources, holding them a little too long isn't a
problem. Consider file handles, for example. OK, you don't want to leak
filehandles anymore than you want to leak memory, opening them and
forgetting to ever close them. But if a file handle stays open a little
longer than you really need it to be, it's not a catastrophe. There are
generally enough spare handles that you can still open new files before
all the old files get closed. So the lag isn't critical.
>
> > And for that, I think representing
> > holding the lock as the lifetime of an object to be a bad idea. Were Caml
> > interested more in doing multithreaded code, I'd recommend something like
> > Java's synchronize keyword.
>
> What's the difference with respect to locking a Mutex?
>
Java explicitly binds the time you hold a mutex to a certain code block-
literally, a set of curly braces {} defines the length of time when you
hold a lock. When you leave that code block, via any means, the lock is
automatically released. So you have code like:
synchronized(foo) {
/* All the code in here has foo locked */
}
/* Code here has foo unlocked */
This is usefull because the programmer doesn't have to remember to unlock
the mutex- the simple act of leaving the block automatically unlocks the
mutex. So:
synchronized(foo) {
if (error) throw AnException();
}
isn't an error. The throw will unlock the mutex on leaving the code
block. This does slow down exceptions a little- now you need to have a
stack of locks to unlock as you throw the exception. However this is way
simpler to implement, and way faster, than having to keep track of all
objects (and their types) on the stack and call the appropriate
destructors as you unwind the stack. And dealing with the corner cases
implied (what happens when you throw an exception, which calls a
destructor for an object on the stack, and that destructor throws another
exception?).
This is the behavior I think was wanted- to bind the mutex lock to a
specific peice of code. I'd much rather do it explicitly, with a language
extension, than a kludge which causes more problems than it solves.
Speaking of which, I've been noodling around with wether it's possible to
implement synchronization automatically in the common case. What objects
need to have synchronization is easy to determine- those objects which can
be reached from 2 or more threads need synchronization. The question is,
what synchronization. Or how much. Can you protect a whole slew of
objects with a single lock?
Obviously, removing as much synchronization as possible is good for
performance. Also, it's possible to construct examples where if multiple
objects are protected by a single lock you don't have deadlock, while if
the objects are each protected by their own lock deadlock is possible.
Brian
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