Messages in this thread | | | Date | Mon, 19 Sep 2022 11:31:07 +0200 | Subject | Re: Correlation CMA size and FORCE_MAX_ZONEORDER | From | David Hildenbrand <> |
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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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