Hi Boris,

On 05/02/2014 10:48, Boris BREZILLON wrote:
> The parent dependency check is only available on the first parent of a given
> clk.
> 
> Add support for strict dependency check: all parents of a given clk must be
> initialized.
> 
> Signed-off-by: Boris BREZILLON <[email protected]>
> ---
> 
> Hello Gregory,
> 
> This patch adds support for strict check on clk dependencies (check if all
> parents specified by an DT clk node are initialized).
> 
> I'm not sure this is what you were expecting (maybe testing the first parent
> is what you really want), so please feel free to tell me if I'm wrong.
> 
> Best Regards,
> 
> Boris
> 
>  drivers/clk/clk.c |   27 +++++++++++++++++++++------
>  1 file changed, 21 insertions(+), 6 deletions(-)
> 
> diff --git a/drivers/clk/clk.c b/drivers/clk/clk.c
> index beb0f8b..6849769 100644
> --- a/drivers/clk/clk.c
> +++ b/drivers/clk/clk.c
> @@ -2543,22 +2543,37 @@ static int parent_ready(struct device_node *np)
>  {
>       struct of_phandle_args clkspec;
>       struct of_clk_provider *provider;
> +     int num_parents;
> +     bool found;
> +     int i;
>  
>       /*
>        * If there is no clock parent, no need to wait for them, then
>        * we can consider their absence as being ready
>        */
> -     if (of_parse_phandle_with_args(np, "clocks", "#clock-cells", 0,
> -                                     &clkspec))
> +     num_parents = of_count_phandle_with_args(np, "clocks", "#clock-cells");
> +     if (num_parents <= 0)
>               return 1;
>  
> -     /* Check if we have such a provider in our array */
> -     list_for_each_entry(provider, &of_clk_providers, link) {
> -             if (provider->node == clkspec.np)
> +     for (i = 0; i < num_parents; i++) {
> +             if (of_parse_phandle_with_args(np, "clocks", "#clock-cells", i,
> +                                            &clkspec))
>                       return 1;
> +
> +             /* Check if we have such a provider in our array */
> +             found = false;
> +             list_for_each_entry(provider, &of_clk_providers, link) {
> +                     if (provider->node == clkspec.np) {
> +                             found = true;
> +                             break;

Hum this means that as soon as you have one parent then you consider it
as ready. It is better of what I have done because I only test the 1st
parent. However I wondered if we should go further by ensuring all the
parents are ready.

If I am right, there is more than one parent only for the muxer. In this
case is it really expected that all the parent are ready?

Thanks,

Gregory

> +                     }
> +             }
> +
> +             if (!found)
> +                     return 0;
>       }
>  
> -     return 0;
> +     return 1;
>  }
>  
>  /**
> 


-- 
Gregory Clement, Free Electrons
Kernel, drivers, real-time and embedded Linux
development, consulting, training and support.
http://free-electrons.com
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