Am 15.03.2012 07:10, schrieb Julia Lawall:
>
>
> On Thu, 15 Mar 2012, Guan Xuetao wrote:
>
>> On Wed, 2012-03-14 at 10:23 +0100, Julia Lawall wrote:
>>>
>>> On Wed, 14 Mar 2012, Guan Xuetao wrote:
>>>
>>>> On Wed, 2012-03-14 at 11:19 +0300, Dan Carpenter wrote:
>>>>> On Wed, Mar 14, 2012 at 04:07:24PM +0800, Guan Xuetao wrote:
>>>>>> puv3_init_dma() is called ONCE when initializing.
>>>>>> In logical, if request_irq(IRQ_DMAERR, *) failed,
>>>>>> free_irq(IRQ_DMA, *)
>>>>>> is unnecessary, and dma device/driver can keep on working.
>>>>>> The patch could be:
>>>>>> ret = request_irq(IRQ_DMAERR, dma_err_handler, 0, "DMAERR",
>>>>>> NULL);
>>>>>> if (ret) {
>>>>>> printk(KERN_CRIT "Can't register IRQ for DMAERR\n");
>>>>>> - free_irq(IRQ_DMA, "DMA");
>>>>>> return ret;
>>>>>> }
>>>>>
>>>>> It seems like you should remove the error return as well?
>>>>>
>>>>> regards,
>>>>> dan carpenter
>>>>>
>>>> The error return value will only generate an extra warning message, and
>>>> have no side-effect.
>>>
>>> The whole thing seems a little strange. I guess your point is that the
>>> call site never looks at the return value? Wouldn't it be better to
>>> make
>>> there be no return value in that case? If there is a return value, some
>>> calling context in the future might take that into account and then the
>>> lack of a free_irq would be a memory leak. Also if the first
>>> request_irq
>>> can never fail, perhaps that should be made explicit by not testing the
>>> return value?
>>>
>>> julia
>> This function is an init_call, not a probe function, and it is only
>> called ONCE.
>> The dma device here has two interrupts, one IRQ_DMA, another IRQ_DMAERR.
>> And the device could work without IRQ_DMAERR.
>> The return value should indicate whether there is something wrong during
>> initialization, so the function needs return errno when any request_irq
>> is failed.
>> For the first request_irq, some code has prepared its resources before
>> this call, so I suppose it successful. However, the return value must be
>> tested.
>
> OK, thank you for the explanation. I will change the patch.
>
hi Julia,
would you mind to add the explaination to the code ? there is a good chance
that someone will find the same problem again.
re,
wh
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