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SubjectRe: [PATCH 1/1] x86: Change _node_to_cpumask_ptr to return const ptr
Vegard Nossum wrote:
> Hi,
> On Tue, Jul 8, 2008 at 7:06 PM, Mike Travis <> wrote:
>>> (v3 is applied already so Mike please send a delta to v3.)
>>> Ingo
>> Subject: [PATCH 1/1] x86: Change _node_to_cpumask_ptr to return const ptr
>> * Strengthen the return type for the _node_to_cpumask_ptr to be
>> a const pointer. This adds compiler checking to insure that
>> node_to_cpumask_map[] is not changed inadvertently.
>> Applies to tip/master with the following patch applied:
>> "[PATCH 1/1] x86: Add check for node passed to node_to_cpumask V3"
>> Signed-off-by: Mike Travis <>
>> ---
>> Note: I did not change node_to_cpumask_ptr() in include/asm-generic/topology.h
>> as node_to_cpumask_ptr_next() does change the cpumask value.
> Hmmm. Does it really?
> #define node_to_cpumask_ptr_next(v, node) \
> _##v = node_to_cpumask(node)
> This doesn't seem to modify it?

Well I thought about it. The pointer (*v) does not change
but the underlying cpumask variable is updated with the
cpumask for the (supposedly) new node number. You can see
that in this code snippet from kernel/sched.c:

for (i = 1; i < SD_NODES_PER_DOMAIN; i++) {
int next_node = find_next_best_node(node, &used_nodes);

node_to_cpumask_ptr_next(nodemask, next_node);
cpus_or(*span, *span, *nodemask);

In the optimized (x86_64) case, the pointer is simply modified
to point to the new node_to_cpumask_map[node] entry. It remains
a pointer to a const value.

But the non-optimized version replaces the const cpumask value
with the new cpumask value. Isn't this breaking the const

> Also, isn't it unfortunate to have the same function return
> const/non-const depending on your arch/config?

But isn't that exactly what it does? (And in reality, the real
protection happens when there is a node_to_cpumask_map[] present.)

But whichever seems more correct is fine with me... ;-)


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