[PATCH v4 2/3] i2c: iproc: Add Broadcom iProc I2C Driver

[Ray Jui]

On 1/17/2015 12:18 PM, Uwe Kleine-König wrote:
> Hello,
> 
> On Sat, Jan 17, 2015 at 11:58:33AM -0800, Ray Jui wrote:
>> On 1/17/2015 8:01 AM, Uwe Kleine-König wrote:
>>> On Fri, Jan 16, 2015 at 02:09:28PM -0800, Ray Jui wrote:
>>>> On 1/15/2015 12:41 AM, Uwe Kleine-König wrote:
>>>>> On Wed, Jan 14, 2015 at 02:23:32PM -0800, Ray Jui wrote:
>>>>>> +	 */
>>>>>> +	val = 1 << M_CMD_START_BUSY_SHIFT;
>>>>>> +	if (msg->flags & I2C_M_RD) {
>>>>>> +		val |= (M_CMD_PROTOCOL_BLK_RD << M_CMD_PROTOCOL_SHIFT) |
>>>>>> +		       (msg->len << M_CMD_RD_CNT_SHIFT);
>>>>>> +	} else {
>>>>>> +		val |= (M_CMD_PROTOCOL_BLK_WR << M_CMD_PROTOCOL_SHIFT);
>>>>>> +	}
>>>>>> +	writel(val, iproc_i2c->base + M_CMD_OFFSET);
>>>>>> +
>>>>>> +	time_left = wait_for_completion_timeout(&iproc_i2c->done, time_left);
>>>>>
>>>>> When the interrupt fires here after the complete timed out and before
>>>>> you disable the irq you still throw the result away.
>>>> Yes, but then this comes down to the fact that if it has reached the
>>>> point that is determined to be a timeout condition in the driver, one
>>>> should really treat it as timeout error. In a normal condition,
>>>> time_left should never reach zero.
>>> I don't agree here. I'm not sure there is a real technical reason,
>>> though. But still if you're in a "success after timeout already over"
>>> situation it's IMHO better to interpret it as success, not timeout.
>>>
>> The thing is, the interrupt should never fire after
>> wait_for_completion_timeout returns zero here. If it does, then the
>> issue is really that the timeout value set in the driver is probably not
>> long enough. I just checked other I2C drivers. I think the way how
>> timeout is handled here is consistent with other I2C drivers.
> In the presence of Clock stretching there is no (theorethical) upper
> limit for the time needed to transfer a given message, is there? So
> (theoretically) you can never be sure not to interrupt an ongoing
> transfer.
> 
Yes. No theoretical upper limit in the case when clock is stretched by
the slave. But how would adding an additional interrupt completion check
below help? I assume you want the the check to be like the following?

	time_left = wait_for_completion_timeout(&iproc_i2c->done, time_left);

	/* disable all interrupts */
	writel(0, iproc_i2c->base + IE_OFFSET);

	if (!time_left && !completion_done()) {
		dev_err(iproc_i2c->device, "transaction timed out\n");

		/* flush FIFOs */
		val = (1 << M_FIFO_RX_FLUSH_SHIFT) |
		      (1 << M_FIFO_TX_FLUSH_SHIFT);
		writel(val, iproc_i2c->base + M_FIFO_CTRL_OFFSET);
		return -ETIMEDOUT;
	}

Does the above code make sense logically? That is,
wait_for_completion_timeout has timed out, and we are doing an
additional check below to make sure it really has timed out?

> And other drivers doing the same is only an excuse to start similar, but
> not to not improve :-)
> 
>>>>>> +static int bcm_iproc_i2c_remove(struct platform_device *pdev)
>>>>>> +{
>>>>>> +	struct bcm_iproc_i2c_dev *iproc_i2c = platform_get_drvdata(pdev);
>>>>>> +
>>>>>> +	i2c_del_adapter(&iproc_i2c->adapter);
>>>>> You need to free the irq before i2c_del_adapter.
>>>>>
>>>> Yes. Thanks. Change back to use devm_request_irq, and use disable_irq
>>>> here before removing the adapter.
>>> The more lightweight approach is to set your device's irq-enable
>>> register to zero and call synchronize_irq. (For a shared irq calling
>>> disable_irq is even wrong here.)
>>>
>> The fact that IRQF_SHARED flag is not set indicates this is a dedicated
>> IRQ line, so I thought using disable_irq here makes sense. But if both
>> you and Wolfram think masking all I2C interrupts at the block level +
>> synchronize_irq is a better approach, I can change to that. Thanks!
> I don't care much. Using synchronize_irq is the more universal approach
> and so more likely correct for someone copying from your driver.
> 
Sure, more universal approach and a good example for others. It takes
care of both cases of dedicated and shared interrupt. I will make that
change. Thanks.
> Best regards
> Uwe
>