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SubjectRe: Correlation CMA size and FORCE_MAX_ZONEORDER
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On 19.09.22 11:17, Michael Nazzareno Trimarchi wrote:
> Hi David
>
> On Mon, Sep 19, 2022 at 10:38 AM David Hildenbrand <david@redhat.com> wrote:
>>
>> On 15.09.22 23:36, Michael Nazzareno Trimarchi wrote:
>>> Hi all
>>
>> Hi,
>>
>>>
>>> Working on a small device with 128MB of memory and using imx_v6_v7
>>> defconfig I found that CMA_SIZE_MBYTES, CMA_SIZE_PERCENTAGE
>>> are not respected. The calculation done does not allow the requested
>>> size. I think that this should be somehow documented and described but
>>> I did not
>>> find the documentation. Does it work this way?
>>>
>>> With CMA_SIZE of 8MB I need to have FORCE_MAX_ZONEORDER=12 if I have
>>> the default FORCE_MAX_ZONEORDER=14 the min size is 32Mb
>>
>> The underlying constraint is that CMA regions require a certain minimum
>> alignment+size. They cannot be arbitrarily in size.
>>
>> CMA_MIN_ALIGNMENT_BYTES expresses that, and corresponds in upstream
>> kernels to the size of a single pageblock.
>>
>> In previous kernels, it used to be the size of the largest buddy
>> allocation granularity (derived from MAX_ORDER, derived from
>> FORCE_MAX_ZONEORDER).
>>
>> On upstream kernels, the FORCE_MAX_ZONEORDER constraint should no longer
>> apply. On most archs, the minimum alignment+size should be 2 MiB
>> (x86-64, aarch64 with 4k base pages) -- the size of a single pageblock.
>>
>> So far the theory. Are you still running into this limitation on
>> upstream kernels?
>>
>
> I can run 6-rc2 on my board. I test again but according to it, if I
> put 4M as CMA in cma=4M in boot
> parameters, the result is 32Mb of CMA. Apart of that seems that
> process lime tiny membench can not even start
> to mblock memory
>

The CMA alignemnt change went into v5.19. If "cma=4M" still gives you >
4M, can you post /proc/meminfo and the early console output?


--
Thanks,

David / dhildenb

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