G' day all,
The kernel would try to pick the best algorithm for raid6 to compute two
syndromes, generally referred to P and Q at boot time. Part of the algorithm
code was showed bellow from lib/raid6/algos.c:
int __init raid6_select_algo(void)
{
const int disks = (65536/PAGE_SIZE)+2;
const struct raid6_calls *gen_best;
const struct raid6_recov_calls *rec_best;
char *syndromes;
void *dptrs[(65536/PAGE_SIZE)+2];
int i;
for (i = 0; i < disks-2; i++)
dptrs[i] = ((char *)raid6_gfmul) + PAGE_SIZE*i;
/* Normal code - use a 2-page allocation to avoid D$ conflict */
syndromes = (void *) __get_free_pages(GFP_KERNEL, 1);
if (!syndromes) {
pr_err("raid6: Yikes! No memory available.\n");
return -ENOMEM;
}
dptrs[disks-2] = syndromes;
dptrs[disks-1] = syndromes + PAGE_SIZE;
/* select raid gen_syndrome function */
gen_best = raid6_choose_gen(&dptrs, disks);
/* select raid recover functions */
rec_best = raid6_choose_recov()
The data set to use for computing syndromes is gfmul table, it was defined as
"u8 raid6_gfmul[256][256]" and size to be 65536 Bytes or 64KB . From
the code we can see it use gfmul table size and PAGE_SIZE to determine
the disk number. If the PAGE_SIZE is 4K, then the number of disks got
to be 18 and 10 for 8K, 3 for 64K. As we all know, raid6 needs at
least 4 disks.
Could we just define a constantly macro for disks as the test program does in
lib/raid6/test/test.c, not depend on page size and not use gfmul
table as the data source of disks?
Move further, bigger page size like 128K would encounter the same problem.
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