On 10/23, Peter Zijlstra wrote:
>
> On Sun, 2006-10-22 at 19:57 +0400, Oleg Nesterov wrote:
> > Depends on
> > tty-signal-tty-locking.patch
> >
> > ->signal->tty is protected by ->siglock, no need to take the global tty_mutex.
> >
> > Signed-off-by: Oleg Nesterov <[email protected]>
>
> ->siglock protects the value of ->signal->tty, not memory it points to;
Yes. That is why we can't avoid tty_mutex in audit_log_exit() (unless we copy
tty->name to the safe location). But is is bad we take tasklist_lock to access
tsk->signal. This 'tsk' should be 'current', we can use get_current_tty().
Or we can use lock_task_sighand() to be safe.
What do you think?
Oleg.
> however since destroying the tty also means clearing all ->signal->tty
> references, which means taking all ->siglocks, just holding the
> ->siglock around this piece of code looks sufficient indeed.
>
> Acked-by: Peter Zijlstra <[email protected]>
>
> > --- rc2-mm2/fs/proc/array.c~ 2006-10-22 19:28:17.000000000 +0400
> > +++ rc2-mm2/fs/proc/array.c 2006-10-22 19:45:52.000000000 +0400
> > @@ -346,20 +346,13 @@ static int do_task_stat(struct task_stru
> > sigemptyset(&sigcatch);
> > cutime = cstime = utime = stime = cputime_zero;
> >
> > - mutex_lock(&tty_mutex);
> > rcu_read_lock();
> > if (lock_task_sighand(task, &flags)) {
> > struct signal_struct *sig = task->signal;
> > - struct tty_struct *tty = sig->tty;
> >
> > - if (tty) {
> > - /*
> > - * sig->tty is not stable, but tty_mutex
> > - * protects us from release_dev(tty)
> > - */
> > - barrier();
> > - tty_pgrp = tty->pgrp;
> > - tty_nr = new_encode_dev(tty_devnum(tty));
> > + if (sig->tty) {
> > + tty_pgrp = sig->tty->pgrp;
> > + tty_nr = new_encode_dev(tty_devnum(sig->tty));
> > }
> >
> > num_threads = atomic_read(&sig->count);
> > @@ -395,7 +388,6 @@ static int do_task_stat(struct task_stru
> > unlock_task_sighand(task, &flags);
> > }
> > rcu_read_unlock();
> > - mutex_unlock(&tty_mutex);
> >
> > if (!whole || num_threads<2)
> > wchan = get_wchan(task);
> >
>
-
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