On 06/02/2017 11:19 AM, Julia Lawall wrote:
>
>
> On Fri, 2 Jun 2017, Milan P. Gandhi wrote:
>
>> On 06/01/2017 08:32 PM, Dan Carpenter wrote:
>>> On Thu, Jun 01, 2017 at 05:41:06PM +0530, Milan P. Gandhi wrote:
>>>> Simplify the check for return code of fcoe_if_init routine
>>>> in fcoe_init function such that we could eliminate need for
>>>> extra 'out_free' label.
>>>>
>>>> Signed-off-by: Milan P. Gandhi <mgandhi@xxxxxxxxxx>
>>>> ---
>>>> drivers/scsi/fcoe/fcoe.c | 10 ++++------
>>>> 1 file changed, 4 insertions(+), 6 deletions(-)
>>>>
>>>> diff --git a/drivers/scsi/fcoe/fcoe.c b/drivers/scsi/fcoe/fcoe.c
>>>> index ea21e7b..fb2a4c9 100644
>>>> --- a/drivers/scsi/fcoe/fcoe.c
>>>> +++ b/drivers/scsi/fcoe/fcoe.c
>>>> @@ -2523,13 +2523,11 @@ static int __init fcoe_init(void)
>>>> fcoe_dev_setup();
>>>>
>>>> rc = fcoe_if_init();
>>>> - if (rc)
>>>> - goto out_free;
>>>> -
>>>> - mutex_unlock(&fcoe_config_mutex);
>>>> - return 0;
>>>> + if (rc == 0) {
>>>> + mutex_unlock(&fcoe_config_mutex);
>>>> + return 0;
>>>> + }
>>>>
>>>> -out_free:
>>>> mutex_unlock(&fcoe_config_mutex);
>>>
>>> Gar... Stop! No1 Don't do this.
>>>
>>> Do failure handling, not success handling.
>>>
>>> People always think they should get creative with the last if statement
>>> in a function. This leads to spaghetti code and it's confusing. Please
>>> never do this again.
>>>
>>> The original is correct and the new code is bad rubbish code.
>>>
>>> regards,
>>> dan carpenter
>>>
>>>
>>
>> Oops, my bad sir. Will keep this in mind.
>
> Still, does the mutex_unlock really need to be duplicated?
>
> julia
>
Hello Julia,
Thanks for your hint! I have found a better way to remove a need for
duplicate mutex_unlock statement and extra 'out_free' label. Will send
the corrected path for the same.
Many thanks,
Milan.
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