[PATCH 15/17] media: st_rc: Don't stay on an IRQ handler forever

[Mason]

On 13/04/2018 11:25, Mauro Carvalho Chehab wrote:
> Em Fri, 13 Apr 2018 11:15:16 +0200
> Mason <slash.tmp at free.fr> escreveu:
> 
>> On 12/04/2018 17:24, Mauro Carvalho Chehab wrote:
>>
>>> As warned by smatch:
>>> 	drivers/media/rc/st_rc.c:110 st_rc_rx_interrupt() warn: this loop depends on readl() succeeding
>>>
>>> If something goes wrong at readl(), the logic will stay there
>>> inside an IRQ code forever. This is not the nicest thing to
>>> do :-)
>>>
>>> So, add a timeout there, preventing staying inside the IRQ
>>> for more than 10ms.
>>>
>>> Signed-off-by: Mauro Carvalho Chehab <mchehab at s-opensource.com>
>>> ---
>>>  drivers/media/rc/st_rc.c | 16 ++++++++++------
>>>  1 file changed, 10 insertions(+), 6 deletions(-)
>>>
>>> diff --git a/drivers/media/rc/st_rc.c b/drivers/media/rc/st_rc.c
>>> index d2efd7b2c3bc..c855b177103c 100644
>>> --- a/drivers/media/rc/st_rc.c
>>> +++ b/drivers/media/rc/st_rc.c
>>> @@ -96,19 +96,24 @@ static void st_rc_send_lirc_timeout(struct rc_dev *rdev)
>>>  
>>>  static irqreturn_t st_rc_rx_interrupt(int irq, void *data)
>>>  {
>>> +	unsigned long timeout;
>>>  	unsigned int symbol, mark = 0;
>>>  	struct st_rc_device *dev = data;
>>>  	int last_symbol = 0;
>>> -	u32 status;
>>> +	u32 status, int_status;
>>>  	DEFINE_IR_RAW_EVENT(ev);
>>>  
>>>  	if (dev->irq_wake)
>>>  		pm_wakeup_event(dev->dev, 0);
>>>  
>>> -	status  = readl(dev->rx_base + IRB_RX_STATUS);
>>> +	/* FIXME: is 10ms good enough ? */
>>> +	timeout = jiffies +  msecs_to_jiffies(10);
>>> +	do {
>>> +		status  = readl(dev->rx_base + IRB_RX_STATUS);
>>> +		if (!(status & (IRB_FIFO_NOT_EMPTY | IRB_OVERFLOW)))
>>> +			break;
>>>  
>>> -	while (status & (IRB_FIFO_NOT_EMPTY | IRB_OVERFLOW)) {
>>> -		u32 int_status = readl(dev->rx_base + IRB_RX_INT_STATUS);
>>> +		int_status = readl(dev->rx_base + IRB_RX_INT_STATUS);
>>>  		if (unlikely(int_status & IRB_RX_OVERRUN_INT)) {
>>>  			/* discard the entire collection in case of errors!  */
>>>  			ir_raw_event_reset(dev->rdev);
>>> @@ -148,8 +153,7 @@ static irqreturn_t st_rc_rx_interrupt(int irq, void *data)
>>>  
>>>  		}
>>>  		last_symbol = 0;
>>> -		status  = readl(dev->rx_base + IRB_RX_STATUS);
>>> -	}
>>> +	} while (time_is_after_jiffies(timeout));
>>>  
>>>  	writel(IRB_RX_INTS, dev->rx_base + IRB_RX_INT_CLEAR);
>>>    
>>
>> Isn't this a place where the iopoll.h helpers might be useful?
>>
>> e.g. readl_poll_timeout()
>>
>> https://elixir.bootlin.com/linux/latest/source/include/linux/iopoll.h#L114
> 
> That won't work. Internally[1], readx_poll_timeout() calls
> usleep_range().
> 
> [1] https://elixir.bootlin.com/linux/latest/source/include/linux/iopoll.h#L43
> 
> It can't be called here, as this loop happens at the irq
> handler.

Sorry, I meant readl_poll_timeout_atomic()

But it might have to be open-coded because of the check for overruns.

Regards.