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SubjectRe: [PATCH 02/10] x86: Change size of node ids from u8 to u16 V3
Mike Travis a écrit :
>> Another point: you want this change, sorry if my previous mail was not detailed enough :
>>
>> --- a/arch/x86/mm/numa_64.c
>> +++ b/arch/x86/mm/numa_64.c
>> @@ -78,7 +78,7 @@ static int __init allocate_cachealigned_memnodemap(void)
>> unsigned long pad, pad_addr;
>>
>> memnodemap = memnode.embedded_map;
>> - if (memnodemapsize <= 48)
>> + if (memnodemapsize <= ARRAY_SIZE(memnode.embedded_map))
>> return 0;
>>
>> pad = L1_CACHE_BYTES - 1;
>>
>
> Hi Eric,
>
> I'm still getting this message with the numa=fake=4 start option:
>
> Faking node 0 at 0000000000000000-0000000028000000 (640MB)
> Faking node 1 at 0000000028000000-0000000050000000 (640MB)
> Faking node 2 at 0000000050000000-0000000078000000 (640MB)
> Faking node 3 at 0000000078000000-000000009ff00000 (639MB)
>
> NUMA: Using 27 for the hash shift.
> Your memory is not aligned you need to rebuild your kernel
> with a bigger NODEMAPSIZE shift=27 No NUMA hash function found.
> NUMA emulation disabled.
>
> Is there something else I need to change? (This is on an AMD box.)
>

Sure, check populate_memnodemap() in arch/x86/mm/numa_64.c


--- a/arch/x86/mm/numa_64.c
+++ b/arch/x86/mm/numa_64.c
@@ -54,7 +54,7 @@ populate_memnodemap(const struct bootnode *nodes, int
numnodes, int shift)
int res = -1;
unsigned long addr, end;

- memset(memnodemap, 0xff, memnodemapsize);
+ memset(memnodemap, 0xff, sizeof(u16)*memnodemapsize);
for (i = 0; i < numnodes; i++) {
addr = nodes[i].start;
end = nodes[i].end;
@@ -63,7 +63,7 @@ populate_memnodemap(const struct bootnode *nodes, int
numnodes, int shift)
if ((end >> shift) >= memnodemapsize)
return 0;
do {
- if (memnodemap[addr >> shift] != 0xff)
+ if (memnodemap[addr >> shift] != NUMA_NO_NODE)
return -1;
memnodemap[addr >> shift] = i;
addr += (1UL << shift);
--
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