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SubjectRe: [NUMA , x86_64] Why memnode_shift is chosen with the lowest possible value ?
Andi Kleen a écrit :
> On Friday 30 September 2005 11:09, Eric Dumazet wrote:
>
>>+ while (populate_memnodemap(nodes, numnodes, shift + 1) >= 0)
>>+ shift++;
>
>
>
> Why shift+1 here?

Thank you Andi fo r reviewing this stuff

The idea it to find the highest shift value, and to break the loop as soon as
the (shift + 1) value gives us an "shift too big" error.

Maybe you want to write :

while (populate_memnodemap(nodes, numnodes, ++shift) >= 0) ;
shift--;
Well, thats only style...


>
>
>>+ if ((end >> shift) >= NODEMAPSIZE)
>>+ return 0;
>
>
> This should be >, not >= shouldn't it?

Let's take an example

end = 0xffffffff;
start = 0xfff00000;
shift = 20
Suppose that NODEMAPSIZE == (end >> shift) == 0xfff
If the test is changed to :

if ((end >> shift) > NODEMAPSIZE)
return 0;
We could do one of the iteration with (addr < end) but (addr >> shift) ==
NODEMAPSIZE

if (memnodemap[NODEMAPSIZE] != 0xff)
return -1;
memnodemap[NODMAPSIZE] = i;
Thats bound violation of memnodemap[]

AFAIK, I wonder why NODEMAPSIZE is 0xfff and not 0x1000, because this off by
one make half of memnodemap[] to be unused for power of two ram size.


>
> -Andi
>
> P.S.: Please cc x86-64 patches to discuss@x86-64.org

Ah thank you

Eric
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