linux-kernel - Re: linux-next: Tree for June 5

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Date:	Fri, 6 Jun 2008 15:50:01 +0200
From:	"Vegard Nossum" <vegard.nossum@...il.com>
To:	"Mike Travis" <travis@....com>
Cc:	"Ingo Molnar" <mingo@...e.hu>,
	"Andrew Morton" <akpm@...ux-foundation.org>,
	"Stephen Rothwell" <sfr@...b.auug.org.au>,
	linux-next@...r.kernel.org, LKML <linux-kernel@...r.kernel.org>
Subject: Re: linux-next: Tree for June 5

On Fri, Jun 6, 2008 at 3:33 PM, Mike Travis <travis@....com> wrote:
> Vegard Nossum wrote:
>>
>> I reproced it with gc 4.1.2. I think the error is somewhere in kernel/sched.c.
>>
>> static int __build_sched_domains(const cpumask_t *cpu_map,
>>                                  struct sched_domain_attr *attr)
>> {
>> ...
>>         for (i = 0; i < MAX_NUMNODES; i++) {
>> ...
>>                 sg = kmalloc_node(sizeof(struct sched_group), GFP_KERNEL, i);
>> ...
>>
>> This code is calling into the allocator with a spurious value of i,
>> which causes SLAB to use an index (of 4 in my case) that is out of
>> bounds for its nodelist array (at least it hasn't been initialized).
>>
>> This bit of code (a bit further down, inside the same loop) is also dubious:
>>
>>                         sg = kmalloc_node(sizeof(struct sched_group),
>>                                           GFP_KERNEL, i);
>>                         if (!sg) {
>>                                 printk(KERN_WARNING
>>                                 "Can not alloc domain group for node %d\n", j);
>>                                 goto error;
>>                         }
>>
>> Where it passes i to kmalloc_node() but reports an allocation for node
>> j. Which one is correct?
>>

Hm, I think I'm wrong and the code is correct. However...

>> Hope this helps, will send an update if I find out more.
>>
>>
>> Vegard
>>
>
> Thanks Vegard for tracking this down.  My thoughts were along the same
> wavelength... ;-)

I applied this patch
@@ -7133,6 +7133,14 @@ static int __build_sched_domains(const
cpumask_t *cpu_map,
                cpus_clear(*covered);

                cpus_and(*nodemask, *nodemask, *cpu_map);
+
+               printk("node %d\n", i);
+               for (j = 0; j < NR_CPUS; ++j)
+                       printk("%c", cpu_isset(j, *nodemask) ? 'X' : '.');
+               printk("\n");
+
+               printk("empty = %d\n", cpus_empty(*nodemask));
+
                if (cpus_empty(*nodemask)) {
                        sched_group_nodes[i] = NULL;
                        continue;

and it shows some really strange output, maybe it makes sense to you:

(the X means cpu is in the node)

Total of 2 processors activated (11976.24 BogoMIPS).
node 0
XX..............................................................................
................................................................................
................................................................................
...............
empty = 0
node 1
XX..............................................................................
................................................................................
................................................................................
...............
empty = 0
l3 = cachep->nodelists[0] (size-64) = ffff81003f824340
node 2
................................................................................
................................................................................
................................................................................
...............
empty = 1
node 3
................................................................................
................................................................................
................................................................................
...............
empty = 1
node 4
X...............................................................................
................................................................................
................................................................................
...............
empty = 0

This is a P4 3.0GHz with 1 physical CPU (but HT, so two logical CPUs).
Yet node 4 is claimed to have a cpu too. That's bogus!

(But I don't think it's an error in sched.c any more, probably the
code that sets up the node maps.)


Vegard

-- 
"The animistic metaphor of the bug that maliciously sneaked in while
the programmer was not looking is intellectually dishonest as it
disguises that the error is the programmer's own creation."
	-- E. W. Dijkstra, EWD1036
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