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>

1 files changed, 4 insertions, 4 deletions

diff --git a/drivers/edac/amd64_edac.c b/drivers/edac/amd64_edac.c
index 9ea1bb4a51e3..f3f4ee909ef5 100644
--- a/drivers/edac/amd64_edac.c
+++ b/drivers/edac/amd64_edac.c
@@ -1406,10 +1406,10 @@ static int f10_match_to_this_node(struct amd64_pvt *pvt, int range,
         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) &&