On Mon, Sep 11, 2023 at 12:39:36PM -0700, Stephen Boyd wrote:
> It's possible for interrupts to get significantly delayed to the point
> that callers of intel_scu_ipc_dev_command() and friends can call the
> function once, hit a timeout, and call it again while the interrupt
> still hasn't been processed. This driver will get seriously confused if
> the interrupt is finally processed after the second IPC has been sent
> with ipc_command(). It won't know which IPC has been completed. This
> could be quite disastrous if calling code assumes something has happened
> upon return from intel_scu_ipc_dev_simple_command() when it actually
> hasn't.
>
> Let's avoid this scenario by simply returning -EBUSY in this case.
> Hopefully higher layers will know to back off or fail gracefully when
> this happens. It's all highly unlikely anyway, but it's better to be
> correct here as we have no way to know which IPC the status register is
> telling us about if we send a second IPC while the previous IPC is still
> processing.
...
> +static struct intel_scu_ipc_dev *intel_scu_ipc_get(struct intel_scu_ipc_dev *scu)
> +{
> + u8 status;
> + if (!scu)
> + scu = ipcdev;
I would write this as
scu = scu ?: ipcdev;
> + if (!scu)
> + return ERR_PTR(-ENODEV);
> +
> + status = ipc_read_status(scu);
> + if (status & IPC_STATUS_BUSY) {
> + dev_dbg(&scu->dev, "device is busy\n");
> + return ERR_PTR(-EBUSY);
> + }
> +
> + return scu;
> +}
Reviewed-by: Andy Shevchenko <andriy.shevchenko@xxxxxxxxxxxxxxx>
--
With Best Regards,
Andy Shevchenko
