Re: [PATCH v7 3/5] clk: Supply the critical clock {init, enable, disable} framework

Lee Jones
2015-07-28T13:01:05Z

On Tue, 28 Jul 2015, Maxime Ripard wrote:

> On Mon, Jul 27, 2015 at 09:53:38AM +0100, Lee Jones wrote:
> > On Mon, 27 Jul 2015, Maxime Ripard wrote:
> >
> > > On Wed, Jul 22, 2015 at 02:04:13PM +0100, Lee Jones wrote:
> > > > These new API calls will firstly provide a mechanisms to tag a clock as
> > > > critical and secondly allow any knowledgeable driver to (un)gate clocks,
> > > > even if they are marked as critical.
> > > >
> > > > Suggested-by: Maxime Ripard <maxime.ripard@free-electrons.com>
> > > > Signed-off-by: Lee Jones <lee.jones@linaro.org>
> > > > ---
> > > > drivers/clk/clk.c | 45 ++++++++++++++++++++++++++++++++++++++++++++
> > > > include/linux/clk-provider.h | 2 ++
> > > > include/linux/clk.h | 30 +++++++++++++++++++++++++++++
> > > > 3 files changed, 77 insertions(+)
> > > >
> > > > diff --git a/drivers/clk/clk.c b/drivers/clk/clk.c
> > > > index 61c3fc5..486b1da 100644
> > > > --- a/drivers/clk/clk.c
> > > > +++ b/drivers/clk/clk.c
> > > > @@ -46,6 +46,21 @@ static struct clk_core *clk_core_lookup(const char *name);
> > > >
> > > > /*** private data structures ***/
> > > >
> > > > +/**
> > > > + * struct critical - Provides 'play' over critical clocks. A clock can be
> > > > + * marked as critical, meaning that it should not be
> > > > + * disabled. However, if a driver which is aware of the
> > > > + * critical behaviour wants to control it, it can do so
> > > > + * using clk_enable_critical() and clk_disable_critical().
> > > > + *
> > > > + * @enabled Is clock critical? Once set, doesn't change
> > > > + * @leave_on Self explanatory. Can be disabled by knowledgeable drivers
> > > > + */
> > > > +struct critical {
> > > > + bool enabled;
> > > > + bool leave_on;
> > > > +};
> > > > +
> > > > struct clk_core {
> > > > const char *name;
> > > > const struct clk_ops *ops;
> > > > @@ -75,6 +90,7 @@ struct clk_core {
> > > > struct dentry *dentry;
> > > > #endif
> > > > struct kref ref;
> > > > + struct critical critical;
> > > > };
> > > >
> > > > struct clk {
> > > > @@ -995,6 +1011,10 @@ static void clk_core_disable(struct clk_core *clk)
> > > > if (WARN_ON(clk->enable_count == 0))
> > > > return;
> > > >
> > > > + /* Refuse to turn off a critical clock */
> > > > + if (clk->enable_count == 1 && clk->critical.leave_on)
> > > > + return;
> > > > +
> > >
> > > I think it should be handled by a separate counting. Otherwise, if you
> > > have two users that marked the clock as critical, and then one of them
> > > disable it...
> > >
> > > > if (--clk->enable_count > 0)
> > > > return;
> > > >
> > > > @@ -1037,6 +1057,13 @@ void clk_disable(struct clk *clk)
> > > > }
> > > > EXPORT_SYMBOL_GPL(clk_disable);
> > > >
> > > > +void clk_disable_critical(struct clk *clk)
> > > > +{
> > > > + clk->core->critical.leave_on = false;
> > >
> > > .. you just lost the fact that it was critical in the first place.
> >
> > I thought about both of these points, which is why I came up with this
> > strategy.
> >
> > Any device which uses the *_critical() API should a) have knowledge of
> > what happens when a particular critical clock is gated and b) have
> > thought about the consequences.
>
> Indeed.
>
> > I don't think we can use reference counting, because we'd need as
> > many critical clock owners as there are critical clocks.
>
> Which we can have if we replace the call to clk_prepare_enable you add
> in your fourth patch in __set_critical_clocks.

What should it be replaced with?

> > Cast your mind back to the reasons for this critical clock API. One
> > of the most important intentions of this API is the requirement
> > mitigation for each of the critical clocks to have an owner
> > (driver).
> >
> > With regards to your second point, that's what 'critical.enabled'
> > is for. Take a look at clk_enable_critical().
>
> I don't think this addresses the issue, if you just throw more
> customers at it, the issue remain with your implementation.
>
> If you have three customers that used the critical API, and if on of
> these calls clk_disable_critical, you're losing leave_on.

That's the idea. See my point above, the one you replied "Indeed"
to. So when a driver uses clk_disable_critical() it's saying, "I know
why this clock is a critical clock, and I know that nothing terrible
will happen if I disable it, as I have that covered". So then if it's
not the last user to call clk_disable(), the last one out the door
will be allowed to finally gate the clock, regardless whether it's
critical aware or not.

Then, when we come to enable the clock again, the critical aware user
then re-marks the clock as leave_on, so not critical un-aware user can
take the final reference and disable the clock.

> Which means that if there's one of the two users left that calls
> clk_disable on it, the clock will actually be disabled, which is
> clearly not what we want to do, as we have still a user that want the
> clock to be enabled.

That's not what happens (at least it shouldn't if I've coded it up
right). The API _still_ requires all of the users to give-up their
reference.

> It would be much more robust to have another count for the critical
> stuff, initialised to one by the __set_critical_clocks function.

If I understand you correctly, we already have a count. We use the
original reference count. No need for one of our own.

Using your RAM Clock (Clock 4) as an example
--------------------------------------------

Early start-up:
Clock 4 is marked as critical and a reference is taken (ref == 1)

Driver probe:
SPI enables Clock 4 (ref == 2)
I2C enables Clock 4 (ref == 3)

Suspend (without RAM driver's permission):
SPI disables Clock 4 (ref == 2)
I2C disables Clock 4 (ref == 1)
/*
* Clock won't be gated because:
* .leave_on is True - can't dec final reference
*/

Suspend (with RAM driver's permission):
/* Order is unimportant */
SPI disables Clock 4 (ref == 2)
RAM disables Clock 4 (ref == 1) /* Won't turn off here (ref > 0)
I2C disables Clock 4 (ref == 0) /* (.leave_on == False) last ref can be taken */
/*
* Clock will be gated because:
* .leave_on is False, so (ref == 0)
*/

Resume:
/* Order is unimportant */
SPI enables Clock 4 (ref == 1)
RAM enables Clock 4 and re-enables .leave_on (ref == 2)
I2C enables Clock 4 (ref == 3)

Hopefully that clears things up.

Please tell me if the code doesn't reflect this strategy, or if you
can see anything wrong with how it operates.

--
Lee Jones
Linaro STMicroelectronics Landing Team Lead
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