On 6/8/23 16:24, kirill.shutemov@xxxxxxxxxxxxxxx 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.
