Re: [PATCH] usertools: fix bind failure from dpdk to kernel

["lihuisong (C)" <lihuisong@huawei.com>]

在 2022/8/10 21:49, Stephen Hemminger 写道:
> On Wed, 10 Aug 2022 08:59:15 +0300
> Krzysztof Kozlowski <krzysztof.kozlowski@linaro.org> wrote:
>
>> On 09/08/2022 14:44, lihuisong (C) wrote:
>>> 在 2022/8/5 23:35, Stephen Hemminger 写道:
>>>> On Fri, 5 Aug 2022 11:10:22 +0800
>>>> Huisong Li <lihuisong@huawei.com> wrote:
>>>>   
>>>>> Currently, the steps for binding device from dpdk driver to kernel
>>>>> driver is as follows:
>>>>> echo $BDF > /sys/bus/pci/drivers/vfio-pci/unbind
>>>>> echo $BDF > /sys/bus/pci/drivers/$kernel_driver/bind
>>>>>
>>>>> This steps cannot bind device from dpdk driver to kernel driver on
>>>>> platform with kernel 5.19. The 'driver_override' must be specify
>>>>> kernel driver before binding device to kernel driver.
>>>>>
>>>>> Fixes: 720b7a058260 ("usertools: fix device binding with kernel tools")
>>>>> Cc: stable@dpdk.org
>>>>>
>>>>> Signed-off-by: Huisong Li <lihuisong@huawei.com>
>>>> Not sure exactly what you did and why.
>>>> The patch seems to just remove the check that the driver
>>>> is in the set of dpdk_drivers.
>>>> .
>>> Currently, the end of the operation binding device from kernel driver to
>>> dpdk driver write '\00' to driver_override file so as to this device can
>>> be bound to any other driver.
>> This could have work but this was not the way to use the
>> driver_override. The kernel ABI document clearly states:
>> "and  may be cleared with an empty string (echo > driver_override)."
>> Documentation/ABI/testing/sysfs-bus-pci
>>
>> Please use the kernel ABI how it is described. Using it in wrong way
>> might sometimes work, sometimes not.
> No, the kernel ABI is what ever worked before.
> The documentation is not the definitive standard.
> .

Before:
Writing "\0" or "\n" to driver_override file can clear
'driver_override' in the pci_dev.

Now(5.19 kernel):
Only "\n" can clear 'driver_override'.
Note: for 'echo > driver_override', the 'buf'
in driver_override_store() is equal to "\n".