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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>
This commit is contained in:
Borislav Petkov 2011-01-13 14:57:34 +01:00
parent 95b0ef55cd
commit e2f79dbdfb
1 changed files with 4 additions and 4 deletions
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8
drivers/edac/amd64_edac.c
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@ -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) &&