Commit 20e4d8139319 ("blk-mq: simplify queue mapping & schedule with each possisble CPU") changed blk_mq_map_swqueue() to iterate over all possible CPUs instead of only present CPUs, but left the comment claiming that non-present CPUs are mapped to the first hctx. The function now uses the tag set's CPU-to-hardware-queue maps for all possible CPUs, without checking whether a CPU is present. Non-present CPUs are therefore not necessarily mapped to hctx0. The fallback to hctx0 in this function handles tag allocation failure, independently of CPU presence. Remove the outdated statement and retain the description of the mapping loop. Signed-off-by: Lei Chen --- block/blk-mq.c | 6 +----- 1 file changed, 1 insertion(+), 5 deletions(-) diff --git a/block/blk-mq.c b/block/blk-mq.c index a26a11c73ee3..59452b855afe 100644 --- a/block/blk-mq.c +++ b/block/blk-mq.c @@ -4173,11 +4173,7 @@ static void blk_mq_map_swqueue(struct request_queue *q) hctx->dispatch_from = NULL; } - /* - * Map software to hardware queues. - * - * If the cpu isn't present, the cpu is mapped to first hctx. - */ + /* Map software to hardware queues. */ for_each_possible_cpu(i) { ctx = per_cpu_ptr(q->queue_ctx, i); -- 2.43.0