Messages in this thread | | | Date | Thu, 8 Jun 2023 16:43:43 -0700 | Subject | Re: [PATCH v11 12/20] x86/virt/tdx: Allocate and set up PAMTs for TDMRs | From | Dave Hansen <> |
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On 6/8/23 16:24, kirill.shutemov@linux.intel.com wrote: >> ret = -EINVAL; >> + if (ret) >> + tdmrs_free_pamt_all(&tdx_tdmr_list); >> + else >> + pr_info("%lu KBs allocated for PAMT.\n", >> + tdmrs_count_pamt_pages(&tdx_tdmr_list) * 4); > "* 4"? This is very cryptic. procfs uses "<< (PAGE_SHIFT - 10)" which > slightly less magic to me. And just make the helper that returns kilobytes > to begin with, if it is the only caller.
Let's look at where this data comes from:
+static unsigned long tdmrs_count_pamt_pages(struct tdmr_info_list *tdmr_list) +{ + unsigned long pamt_npages = 0; + int i; + + for (i = 0; i < tdmr_list->nr_consumed_tdmrs; i++) { + unsigned long pfn, npages; + + tdmr_get_pamt(tdmr_entry(tdmr_list, i), &pfn, &npages); + pamt_npages += npages; + }
OK, so tdmr_get_pamt() is getting it in pages. How is it *stored*?
+static void tdmr_get_pamt(struct tdmr_info *tdmr, unsigned long *pamt_pfn, + unsigned long *pamt_npages) +{ ... + pamt_sz = tdmr->pamt_4k_size + tdmr->pamt_2m_size + tdmr->pamt_1g_size; ++ *pamt_pfn = PHYS_PFN(pamt_base); + *pamt_npages = pamt_sz >> PAGE_SHIFT; +}
Oh, it's actually stored in bytes. So to print it out you actually convert it from bytes->pages->kbytes. Not the best.
If tdmr_get_pamt() just returned 'pamt_size_bytes', you could do one conversion at:
free_contig_range(pamt_pfn, pamt_size_bytes >> PAGE_SIZE);
and since tdmrs_count_pamt_pages() has only one caller you can just make it: tdmrs_count_pamt_kb(). The print becomes:
pr_info("%lu KBs allocated for PAMT.\n", tdmrs_count_pamt_kb(&tdx_tdmr_list) * 4);
and tdmrs_count_pamt_kb() does something super fancy like:
return pamt_size_bytes / 1024;
which makes total complete obvious sense and needs zero explanation.
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