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 9ea1bb4a51e3ceee34ff68a1ea3bd8e9f2378df3..f3f4ee909ef51f8e7ee863ba259e3c2d354d9452 100644 (file)
--- 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) &&