On Tue, 03 Dec 2013 10:28:13 +0800 Zhang Yanfei <zhangyanfei@xxxxxxxxxxxxxx> wrote:
> From: Tang Chen <tangchen@xxxxxxxxxxxxxx>
>
> At very early time, the kernel have to use some memory such as
> loading the kernel image. We cannot prevent this anyway. So any
> node the kernel resides in should be un-hotpluggable.
>
> @@ -555,6 +563,30 @@ static void __init numa_init_array(void)
> }
> }
>
> +static void __init numa_clear_kernel_node_hotplug(void)
> +{
> + int i, nid;
> + nodemask_t numa_kernel_nodes;
> + unsigned long start, end;
> + struct memblock_type *type = &memblock.reserved;
> +
> + /* Mark all kernel nodes. */
> + for (i = 0; i < type->cnt; i++)
> + node_set(type->regions[i].nid, numa_kernel_nodes);
> +
> + /* Clear MEMBLOCK_HOTPLUG flag for memory in kernel nodes. */
> + for (i = 0; i < numa_meminfo.nr_blks; i++) {
> + nid = numa_meminfo.blk[i].nid;
> + if (!node_isset(nid, numa_kernel_nodes))
> + continue;
> +
> + start = numa_meminfo.blk[i].start;
> + end = numa_meminfo.blk[i].end;
> +
> + memblock_clear_hotplug(start, end - start);
> + }
> +}
Shouldn't numa_kernel_nodes be initialized?
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