Re: [RFC 1/2] workqueue: use the nearest NUMA node, not the local one
From: Lai Jiangshan <hidden>
Date: 2014-07-18 08:11:49
Also in:
linux-mm, lkml
Hi, I'm curious about what will it happen when alloc_pages_node(memoryless_node). If the memory is allocated from the most preferable node for the @memoryless_node, why we need to bother and use cpu_to_mem() in the caller site? If not, why the memory allocation subsystem refuses to find a preferable node for @memoryless_node in this case? Does it intend on some purpose or it can't find in some cases? Thanks, Lai Added CC to Tejun (workqueue maintainer). On 07/18/2014 07:09 AM, Nishanth Aravamudan wrote:
quoted hunk ↗ jump to hunk
In the presence of memoryless nodes, the workqueue code incorrectly uses cpu_to_node() to determine what node to prefer memory allocations come from. cpu_to_mem() should be used instead, which will use the nearest NUMA node with memory. Signed-off-by: Nishanth Aravamudan <redacted>diff --git a/kernel/workqueue.c b/kernel/workqueue.c index 35974ac..0bba022 100644 --- a/kernel/workqueue.c +++ b/kernel/workqueue.c@@ -3547,7 +3547,12 @@ static struct worker_pool *get_unbound_pool(const struct workqueue_attrs *attrs) for_each_node(node) { if (cpumask_subset(pool->attrs->cpumask, wq_numa_possible_cpumask[node])) { - pool->node = node; + /* + * We could use local_memory_node(node) here, + * but it is expensive and the following caches + * the same value. + */ + pool->node = cpu_to_mem(cpumask_first(pool->attrs->cpumask)); break; } }@@ -4921,7 +4926,7 @@ static int __init init_workqueues(void) pool->cpu = cpu; cpumask_copy(pool->attrs->cpumask, cpumask_of(cpu)); pool->attrs->nice = std_nice[i++]; - pool->node = cpu_to_node(cpu); + pool->node = cpu_to_mem(cpu); /* alloc pool ID */ mutex_lock(&wq_pool_mutex); --To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/