Re: Killing the tty lock

From: Greg KH
Date: Wed May 02 2012 - 16:36:55 EST


On Wed, May 02, 2012 at 11:45:15AM +0100, Alan Cox wrote:
> > It's mostly pretty "sane", but what is this:
> >
> > > +/*
> > > + * Getting the big tty mutex for a pair of ttys with lock ordering
> > > + * On a non pty/tty pair tty2 can be NULL which is just fine.
> > > + */
> > > +void __lockfunc tty_lock_pair(struct tty_struct *tty,
> > > + struct tty_struct *tty2)
> > > +{
> > > + if (tty < tty2) {
> > > + tty_lock(tty);
> > > + tty_lock(tty2);
> > > + } else {
> > > + if (tty2 && tty2 != tty)
> > > + tty_lock(tty2);
> > > + tty_lock(tty);
> > > + }
> > > +}
> > > +EXPORT_SYMBOL(tty_lock_pair);
> > > +
> > > +void __lockfunc tty_unlock_pair(struct tty_struct *tty,
> > > + struct tty_struct *tty2)
> > > +{
> > > + tty_unlock(tty);
> > > + if (tty2 && tty2 != tty)
> > > + tty_unlock(tty2);
> > > +}
> > > +EXPORT_SYMBOL(tty_unlock_pair);
> >
> > for?
>
> We need to take locks on a pair of tty devices at the same time in some
> cases (pty/tty pairs).

Ok.

> > And what's with the comparing of pointers as "<"? How portable is that
> > really, and how are we supposed to control the memory location of these
> > structures?
>
> You don't need to. The point is that we must lock any arbitrary pair of
> tty structs in a defined order. Pointer comparisons work just fine for
> this. The fs layer uses similar logic for inode locking. We only care
> that for any given pair of objects the lock ordering is consistent.

Ah, ok, that makes more sense, sorry, I didn't understand that.

greg
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