Re: [PATCH 2/4 v5] memstick: Prevent memstick host from getting runtime suspended during card detection

[Ulf Hansson]

On 24 October 2018 at 10:49, Kai-Heng Feng <kai.heng.feng@xxxxxxxxxxxxx> wrote:
> We can use MEMSTICK_POWER_{ON,OFF} along with pm_runtime_{get,put}
> helpers to let memstick host support runtime pm.
>
> There's a small window between memstick_detect_change() and its queued
> work, memstick_check(). In this window the rpm count may go down to zero
> before the memstick host powers on, so the host can be inadvertently
> suspended.
>
> Increment rpm count before calling memstick_check(), and decrement rpm
> count afterward, as now we are sure the memstick host should be
> suspended or not.
>
> Signed-off-by: Kai-Heng Feng <kai.heng.feng@xxxxxxxxxxxxx>
> ---
>  drivers/memstick/core/memstick.c | 4 ++++
>  1 file changed, 4 insertions(+)
>
> diff --git a/drivers/memstick/core/memstick.c b/drivers/memstick/core/memstick.c
> index 76382c858c35..5f16a8826401 100644
> --- a/drivers/memstick/core/memstick.c
> +++ b/drivers/memstick/core/memstick.c
> @@ -18,6 +18,7 @@
>  #include <linux/delay.h>
>  #include <linux/slab.h>
>  #include <linux/module.h>
> +#include <linux/pm_runtime.h>
>
>  #define DRIVER_NAME "memstick"
>
> @@ -209,6 +210,7 @@ static int memstick_dummy_check(struct memstick_dev *card)
>   */
>  void memstick_detect_change(struct memstick_host *host)
>  {
> +       pm_runtime_get_noresume(host->dev.parent);
>         queue_work(workqueue, &host->media_checker);
>  }
>  EXPORT_SYMBOL(memstick_detect_change);
> @@ -479,6 +481,8 @@ static void memstick_check(struct work_struct *work)
>                 host->set_param(host, MEMSTICK_POWER, MEMSTICK_POWER_OFF);
>
>         mutex_unlock(&host->lock);
> +
> +       pm_runtime_put(host->dev.parent);
>         dev_dbg(&host->dev, "memstick_check finished\n");
>  }
>

I am not sure this works, sorry.

More precisely, I don't think there is a guarantee that the calls to
pm_runtime_get|put*() becomes properly balanced. In principle
memstick_detect_change() could be called, without actually causing a
new work to be scheduled if there is already such a work in the queue
(depends on the workqueue configuration). Isn't it so?

Kind regards
Uffe