Re: Semaphore

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On 2012-02-22 11:53:12 (+0000), Anuz Pratap Singh Tomar <chambilkethakur@xxxxxxxxx> wrote:
> On Wed, Feb 22, 2012 at 9:48 AM, Kristof Provost <kristof@xxxxxxxxxx> wrote:
> > On 2012-02-22 11:01:52 (+0200), Konstantin Zertsekel <zertsekel@xxxxxxxxx>
> > wrote:
> > > On Tue, Feb 21, 2012 at 6:14 PM, Dave Hylands <dhylands@xxxxxxxxx>
> > wrote:
> > > > I'm assuming that the semaphore is one which is held across multiple
> > > > calls into the kernel, otherwise you don't have an issue in the first
> > > > place, unless there is a bug on the kernel side of things which
> > > > actually caused the process to terminate.
> > >
> > > Ok, but what happens if things go wrong?
> > > For example, it driver exists abnormally (segmentation fault or
> > something)?
> > > Anyway, it seems very strange that the responsibility is of a driver
> > alone!
> > > There is the *kernel* in the system to take care of abnormal
> > > situation, not the exit function of a driver...
> > >
> > The driver is part of the kernel. If it dies the whole kernel can
> > (perhaps even should) die.
> >
> > There are systems, like Minix, where drivers don't run in kernel mode
> > and where a crashing driver won't take the system down.
> > There are advantages and disadvantages to that approach.
> > See
> >
> >
> I am curious though if userspace gets segmentation fault, which is SIGSEGV,
> kernel should be the one sending that to user space. And while sending
> SIGSEGV, it must be doing some exit cleanup, wherein it frees all
> resources. However unlike memory, i haven't seen exit code which frees lock
> as well?
You're talking about a user space application crashing, Konstantin was
talking about crashes in kernel space. These are two very different

In either case we're talking about a lock inside the kernel.

For the first case:
Imagine a driver which only one app can use at a time (perhaps a serial
port). The kernel will take a lock when the user space app open()s the
device node and release it when it close()s.
If the app segfaults in between the open() and close() the lock will
still get released. The kernel always cleans up open files for stopped
processes, regardless of how they stop. During this cleanup the kernel
will close() the device node, and as a result the driver will release
the lock.

In the second case:
A kernel thread has a lock held, for whatever reason. It can be
terminated for a variety of reasons (*all* of them bugs or hardware
problems). For example it fails a BUG_ON check, or causes a fault by
accessing unmapped memory. In that case the oops handler will terminate
the thread, but in that case the lock will still be held, and never
This results in an incorrect state of the kernel! Some things might
still work, others will be broken.


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