On Mon, May 23, 2022 at 02:34:49PM +0200, Andrew Lunn wrote:
> > --- a/drivers/net/phy/phy_device.c
> > +++ b/drivers/net/phy/phy_device.c
> > @@ -283,8 +283,11 @@ static __maybe_unused int mdio_bus_phy_suspend(struct device *dev)
> > * may call phy routines that try to grab the same lock, and that may
> > * lead to a deadlock.
> > */
> > - if (phydev->attached_dev && phydev->adjust_link)
> > + if (phydev->attached_dev && phydev->adjust_link) {
> > + if (phy_interrupt_is_valid(phydev))
> > + synchronize_irq(phydev->irq);
> > phy_stop_machine(phydev);
> > + }
>
> What is this hunk trying to achieve? As far as i know, interrupts have
> not been disabled. So as soon as the call to synchronize_irq()
> finishes, could well be another interrupt happens.
That other interrupt would bail out of phy_interrupt() because
the is_prepared flag is set on the PHY's struct device, see
first hunk of the patch.
The problem is that an interrupt may occur before the system
sleep transition commences. phy_interrupt() will notice that
is_prepared is not (yet) set, hence invokes drv->handle_interrupt().
Let's say the IRQ thread is preempted at that point, the system
sleep transition is started and mdio_bus_phy_suspend() is run.
It calls phy_stop_machine(), so the state machine is now stopped.
Now phy_interrupt() continues, and the PHY driver's ->handle_interrupt()
callback starts the state machine. Boom, that's not what we want.
So the synchronize_irq() ensures that any already running
phy_interrupt() runs to completion before phy_stop_machine()
is called. It doesn't matter if another interrupt occurs
because then is_prepared will have been set and therefore
phy_interrupt() won't call drv->handle_interrupt().
Let me know if I haven't explained it in sufficient clarity,
I'll be happy to try again. :)
I'm more concerned about the first hunk of the patch because I'm
not sure I got the wakeup stuff right...
Thanks,
Lukas
