Re: [PATCH 4/4] PCI: fix the io resource alignment calculation in pbus_size_io()

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On Tue, Jul 09, 2013 at 11:38:06AM -0600, Bjorn Helgaas wrote:
>On Mon, Jul 8, 2013 at 9:20 PM, Wei Yang <weiyang@xxxxxxxxxxxxxxxxxx> wrote:
>> On Mon, Jul 08, 2013 at 03:15:13PM -0600, Bjorn Helgaas wrote:
>>>On Mon, Jul 1, 2013 at 9:10 AM, Wei Yang <weiyang@xxxxxxxxxxxxxxxxxx> wrote:
>>>> In commit 462d9303 ("PCI: Align P2P windows using pcibios_window_alignment()"),
>>>> it introduce a new method to calculate the window alignment of P2P bridge.
>>>>
>>>> When the io_window_1k is set,  the calculation for the io resource alignment
>>>> is different from the original one. In the original logic before 462d9303,
>>>> the alignment is no bigger than 4K even the io_window_1k is set. The logic
>>>> introduced in 462d9303 will limit the alignment to 1k in this case.
>>>>
>>>> This patch fix this issue.
>>>>
>>>> Signed-off-by: Wei Yang <weiyang@xxxxxxxxxxxxxxxxxx>
>>>> Reviewed-by: Gavin Shan <shangw@xxxxxxxxxxxxxxxxxx>
>>>> Reviewed-by: Ram Pai <linuxram@xxxxxxxxxx>
>>>> ---
>>>>  drivers/pci/setup-bus.c |    4 ++++
>>>>  1 files changed, 4 insertions(+), 0 deletions(-)
>>>>
>>>> diff --git a/drivers/pci/setup-bus.c b/drivers/pci/setup-bus.c
>>>> index bd0ce39d..5c60ca0 100644
>>>> --- a/drivers/pci/setup-bus.c
>>>> +++ b/drivers/pci/setup-bus.c
>>>> @@ -755,6 +755,10 @@ static void pbus_size_io(struct pci_bus *bus, resource_size_t min_size,
>>>>                 return;
>>>>
>>>>         io_align = min_align = window_alignment(bus, IORESOURCE_IO);
>>>> +       /* Don't exceed 4KiB for windows requesting 1KiB alignment */
>>>> +       if (bus->self->io_window_1k && io_align == PCI_P2P_DEFAULT_IO_ALIGN_1K)
>>>> +               io_align = PCI_P2P_DEFAULT_IO_ALIGN;
>>>
>>>Please explain why we need this change, with some actual values that
>>>show the problem.  We need to know what the problem is, not merely
>>>that the code behaves differently than it did before 462d9303.
>>
>> Yep, sorry for not listing the exact problem value.
>>
>> Assume:
>>         1. pcibios_window_alignment() return 1.
>>         2. window_alignment() return PCI_P2P_DEFAULT_IO_ALIGN_1K.
>>         3. one of the child device has an IO resource with size of 2K.
>>
>> Result comparison:
>>
>>                     Before 462d9303             After 462d9303
>>     min_align       1k                          1k
>>                                         |
>>                              after loop |
>>                                         V
>>     min_align       2k                          2k
>>                                         |
>>                          check boundary |
>>                                         V
>>     min_align       2k                          1k
>>
>> In this case, with 462d9303 the min_align will be set back to 1k even one of
>> the child require 2k alignment.
>>
>>>
>>>It appears to me that this change will break the ability to use 1K
>>>windows.  For example, assume a bridge that supports 1K windows.
>>>Assume we're using the default pcibios_window_alignment().  Currently
>>>window_alignment() on the secondary bus returns
>>>PCI_P2P_DEFAULT_IO_ALIGN_1K (0x400, which is 1K), so io_align = 0x400.
>>>
>>>With your change, I think io_align will be bumped back up to 4K in
>>>this case, so we'll lose the ability to allocate a 1K window.
>>
>> After applying the change:
>>
>> Assume:
>>         1. pcibios_window_alignment() return 1.
>>         2. window_alignment() return PCI_P2P_DEFAULT_IO_ALIGN_1K.
>>         3. one of the child device has an IO resource with size of 2K.
>
>What happens if no child has an I/O resource larger than 1K?  Can we
>allocate a 1K window with 1K alignment in that case?
>

Yes, it could. The result comparison would look like this.
Since no child has an I/O resource larger than 1k, the min_align will remain
1k after loop. And because io_align(4K) is larger than min_align(1k), the
final min_align would be 1k.

In this case, the code from commit 462d9303 and my patch both works.

 Result comparison:
                     with  462d9303             with this patch
     min_align       1k                          1k
     io_align        1k                          4k
                                         |
                              after loop |
                                         V
     min_align       1k                          1k
     io_align        1k                          4k
                                         |
                          check boundary |
                                         V
     min_align       1k                          1k
     io_align        1k                          4k

>> Result comparison:
>>
>>                     with  462d9303             with this patch
>>     min_align       1k                          1k
>>     io_align        1k                          4k
>>                                         |
>>                              after loop |
>>                                         V
>>     min_align       2k                          2k
>>     io_align        1k                          4k
>>                                         |
>>                          check boundary |
>>                                         V
>>     min_align       1k                          2k
>>     io_align        1k                          4k
>>
>> With this patch, in the same case as above, the min_align is 2k after
>> calculation.
>>
>> In my mind, the min_align is the lower bound, io_align is the upper bound. The
>> final result of min_align should be in this range.
>>
>> Is my understanding correct? or I missed something important?
>>
>>>
>>>>         list_for_each_entry(dev, &bus->devices, bus_list) {
>>>>                 int i;
>>>>
>>>> --
>>>> 1.7.5.4
>>>>
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>>
>> --
>> Richard Yang
>> Help you, Help me
>>
>--
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-- 
Richard Yang
Help you, Help me

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