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charset=UTF-8 Content-Transfer-Encoding: 7bit On 9/29/25 10:41, Edgecombe, Rick P wrote: > On Tue, 2025-09-23 at 15:45 +0800, Binbin Wu wrote: >>> +/* >>> + * Allocate PAMT reference counters for all physical memory. >>> + * >>> + * It consumes 2MiB for every 1TiB of physical memory. >>> + */ >>> +static int init_pamt_metadata(void) >>> +{ >>> + size_t size = max_pfn / PTRS_PER_PTE * sizeof(*pamt_refcounts); >> Is there guarantee that max_pfn is PTRS_PER_PTE aligned? >> If not, it should be rounded up. > Vmalloc() should handle it? vmalloc() will, for instance, round up to 2 pages if you ask for 4097 bytes in 'size'. But that's not the problem. The 'size' calculation itself is the problem. You need exactly 2 MiB for every 1 TiB of memory, so let's say we have: max_pfn = 1<<28 (where 28 == 40-PAGE_SIZE) then size would be *exactly* 1<<21 (2 MiB). Right? But what if: max_pfn = (1<<28) + 1 Then size needs to be one more page. Right? But what would the code do?