[PATCH v2 1/7] clk: Add a generic clock infrastructure
robherring2 at gmail.com
Mon Oct 3 15:24:52 UTC 2011
On 10/03/2011 09:25 AM, Mark Brown wrote:
> On Mon, Oct 03, 2011 at 09:17:30AM -0500, Rob Herring wrote:
>> On 09/22/2011 05:26 PM, Mike Turquette wrote:
> A lot of stuff that should really have been cut plus...
>>> + if (clk->ops->get_parent)
>>> + /* We don't to lock against prepare/enable here, as
>>> + * the clock is not yet accessible from anywhere */
>>> + clk->parent = clk->ops->get_parent(clk->hw);
>> I don't think this is going to work. This implies that the parent clock
>> is already registered. For simple clk trees, that's probably not an
>> issue, but for chips with lots of muxing it will be impossible to get
>> the order correct for all cases. This is not an issue today as most
>> clocks are statically created.
>> I think what is needed is a 2 stage init. The 1st stage to create all
>> the clocks and a 2nd stage to build the tree once all clocks are created.
>> Tracking the parents using struct clk_hw instead would help as long as
>> clocks are still statically allocated. However, that won't help for
> This isn't in any way specific to clocks, right now the likely solution
> looks to be Grant's changes for retrying probe() as new devices come on
> line. With that devices can return a code from their probe() which
> tells the driver core that they couldn't get all the resources they need
> and that it should retry the probe() if more devices come on-line.
Except SOC clocks are initialized very early before timers are up and
there can be a very high number of dependencies (every clock except
fixed clocks). With the driver probe retry, retrying is the exception,
not the rule.
Retrying would require every caller to maintain a list of clks to
retry. With 2 stages, you can move that into the core clock code.
There are not typically a large number of board-level/driver created
clocks, so ensuring correct register order is not really a problem. In
cases where there is a cross-driver dependency, the probe retry is a
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