Messages in this thread | | | From | "Huang, Kai" <> | Subject | Re: [PATCH v11 12/20] x86/virt/tdx: Allocate and set up PAMTs for TDMRs | Date | Mon, 12 Jun 2023 02:52:13 +0000 |
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On Thu, 2023-06-08 at 16:43 -0700, Dave Hansen wrote: > 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);
I thought making tdmr_get_pamt() return pamt_pfn and pamt_npages would be more clear that PAMTs must be in 4K granularity, but I guess it doesn't matter anyway.
If we return bytes for PAMT size, I think we should also return physical address instead of PFN for PAMT start?
I'll change tdmr_get_pamt() to return physical address and bytes for PAMT location and size respectively. Please let me know if you have any comments.
> > 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.
Will do.
Thanks for the feedback.
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