Re: [NUMA , x86_64] Why memnode_shift is chosen with the lowest possiblevalue ?
From: Eric Dumazet
Date: Tue Oct 04 2005 - 16:12:29 EST
Andi Kleen a Ãcrit :
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@xxxxxxxxxx
Ah thank you
Eric
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