Re: + syscalls-x86-add-__nr_kcmp-syscall-v8.patch added to -mm tree

Cyrill Gorcunov
2012-02-16T15:13:47Z

On Thu, Feb 16, 2012 at 03:49:54PM +0100, Oleg Nesterov wrote:
> On 02/16, Cyrill Gorcunov wrote:
> >
> > +static int access_trylock(struct task_struct *task)
> > +{
> > + if (!mutex_trylock(&task->signal->cred_guard_mutex))
> > + return -EBUSY;
> > +
> > + if (!ptrace_may_access(task, PTRACE_MODE_READ)) {
> > + mutex_unlock(&task->signal->cred_guard_mutex);
> > + return -EPERM;
> > + }
> > +
> > + return 0;
> > +}
>
> OK, this looks correct, but I don't really understand _trylock.
> This means the caller should always retry if -EBUSY, and
> kcmp(pid, pid) can never succeed. Sure, kcmp() doesn't make
> a lot of sense if pid1 == pid2, but this looks a bit strange.
>

Hi Oleg, sure I can make it this way, also I think if pid1 == pid2
and idx1 == idx2 I can return 0 immediately.

> You could simply do
> int mutex_double_lock_killable(struct mutex *m1, struct mutex *m2)
> {
> int err;
>
> if (m2 > m1)
> swap(m1, m2);
>
> err = mutex_lock_killable(m1);
>
> if (!err && likely(m1 != m2)) {
> err = mutex_lock_killable_nested(m2);
> if (err)
> mutex_unlock(m1);
> }
>
> return err;
> }
>
> but I won't insist.

Initially I wanted kcmp would be brining minimum impact
and if mutex is already taken by someone, we would not sleep
but return immediately with -EBUSY and it would be up to caller
to deside if to try again or make some delay first. I simply
not sure what is better here.

Cyrill