amd64_edac: Correct node interleaving removal
When node interleaving is enabled, a subset of the addr[14:12] bits has
to be removed in order to get the normalized DCT address of the DRAM
channel. The actual number of bits to remove is determined by F1x[1,
0][7C:40][IntlvEn]. Do this correctly.
Signed-off-by: Borislav Petkov <borislav.petkov@amd.com>
diff --git a/drivers/edac/amd64_edac.c b/drivers/edac/amd64_edac.c
index 9ea1bb4..f3f4ee9 100644
--- a/drivers/edac/amd64_edac.c
+++ b/drivers/edac/amd64_edac.c
@@ -1406,10 +1406,10 @@
chan_addr = f10_get_norm_dct_addr(pvt, range, sys_addr,
high_range, dct_sel_base);
- /* remove Node ID (in case of node interleaving) */
- tmp = chan_addr & 0xFC0;
-
- chan_addr = ((chan_addr >> hweight8(intlv_en)) & GENMASK(12, 47)) | tmp;
+ /* Remove node interleaving, see F1x120 */
+ if (intlv_en)
+ chan_addr = ((chan_addr >> (12 + hweight8(intlv_en))) << 12) |
+ (chan_addr & 0xfff);
/* remove channel interleave and hash */
if (dct_interleave_enabled(pvt) &&