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

[Cyrill Gorcunov]

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
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