Messages in this thread Patch in this message | | | Date | Tue, 08 Jul 2008 13:51:07 -0700 | From | Mike Travis <> | Subject | Re: [PATCH 1/1] x86: Change _node_to_cpumask_ptr to return const ptr |
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Vegard Nossum wrote: > On Tue, Jul 8, 2008 at 8:05 PM, Mike Travis <travis@sgi.com> wrote: >>>> 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 >> attribute? > > No, I think the pointer really should be const. This doesn't guarantee > that the value doesn't change behind our backs, it only prevents us > from modifying it ourselves. > > > Vegard >
Is this what you had in mind:
--- linux-2.6.tip.orig/include/asm-generic/topology.h +++ linux-2.6.tip/include/asm-generic/topology.h @@ -60,7 +60,7 @@ #ifndef node_to_cpumask_ptr #define node_to_cpumask_ptr(v, node) \ - cpumask_t _##v = node_to_cpumask(node), *v = &_##v + const cpumask_t _##v = node_to_cpumask(node), *v = &_##v #define node_to_cpumask_ptr_next(v, node) \ _##v = node_to_cpumask(node)
(It's taking a while as now I need to do some cross-compile testing.)
Thanks, Mike
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