Messages in this thread | | | Date | Tue, 8 Jul 2008 20:22:57 +0200 | From | "Vegard Nossum" <> | Subject | Re: [PATCH 1/1] x86: Change _node_to_cpumask_ptr to return const ptr |
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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
-- "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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