EDAC/sb_edac: Fix computation of channel address

Large memory Haswell-EX systems with multiple DIMMs per channel were
sometimes reporting the wrong DIMM.

Found three problems:

 1) Debug printouts for socket and channel interleave were not interpreting
    the register fields correctly. The socket interleave field is a 2^X
    value (0=1, 1=2, 2=4, 3=8). The channel interleave is X+1 (0=1, 1=2,
    2=3. 3=4).

 2) Actual use of the socket interleave value didn't interpret as 2^X

 3) Conversion of address to channel address was complicated, and wrong.

Signed-off-by: Tony Luck <tony.luck@intel.com>
Acked-by: Aristeu Rozanski <arozansk@redhat.com>
Cc: Borislav Petkov <bp@alien8.de>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mauro Carvalho Chehab <mchehab@osg.samsung.com>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Cc: linux-edac@vger.kernel.org
Cc: stable@vger.kernel.org
Signed-off-by: Ingo Molnar <mingo@kernel.org>
This commit is contained in:
Luck, Tony 2016-03-09 16:40:48 -08:00 committed by Ingo Molnar
parent 92b0729c34
commit eb1af3b71f

View File

@ -1839,8 +1839,8 @@ static void get_memory_layout(const struct mem_ctl_info *mci)
edac_dbg(0, "TAD#%d: up to %u.%03u GB (0x%016Lx), socket interleave %d, memory interleave %d, TGT: %d, %d, %d, %d, reg=0x%08x\n",
n_tads, gb, (mb*1000)/1024,
((u64)tmp_mb) << 20L,
(u32)TAD_SOCK(reg),
(u32)TAD_CH(reg),
(u32)(1 << TAD_SOCK(reg)),
(u32)TAD_CH(reg) + 1,
(u32)TAD_TGT0(reg),
(u32)TAD_TGT1(reg),
(u32)TAD_TGT2(reg),
@ -2118,7 +2118,7 @@ static int get_memory_error_data(struct mem_ctl_info *mci,
}
ch_way = TAD_CH(reg) + 1;
sck_way = TAD_SOCK(reg) + 1;
sck_way = 1 << TAD_SOCK(reg);
if (ch_way == 3)
idx = addr >> 6;
@ -2175,7 +2175,7 @@ static int get_memory_error_data(struct mem_ctl_info *mci,
n_tads,
addr,
limit,
(u32)TAD_SOCK(reg),
sck_way,
ch_way,
offset,
idx,
@ -2190,18 +2190,12 @@ static int get_memory_error_data(struct mem_ctl_info *mci,
offset, addr);
return -EINVAL;
}
addr -= offset;
/* Store the low bits [0:6] of the addr */
ch_addr = addr & 0x7f;
/* Remove socket wayness and remove 6 bits */
addr >>= 6;
addr = div_u64(addr, sck_xch);
#if 0
/* Divide by channel way */
addr = addr / ch_way;
#endif
/* Recover the last 6 bits */
ch_addr |= addr << 6;
ch_addr = addr - offset;
ch_addr >>= (6 + shiftup);
ch_addr /= ch_way * sck_way;
ch_addr <<= (6 + shiftup);
ch_addr |= addr & ((1 << (6 + shiftup)) - 1);
/*
* Step 3) Decode rank