Re: [PATCH v2 08/11] blk-mq: Add blk_mq_ops.init_request_no_hctx()
From: John Garry <hidden>
Date: 2021-08-18 08:46:37
Also in:
linux-scsi, lkml
On 18/08/2021 08:38, Ming Lei wrote:
On Mon, Aug 09, 2021 at 10:29:35PM +0800, John Garry wrote:quoted
Add a variant of the init_request function which does not pass a hctx_idx arg. This is important for shared sbitmap support, as it needs to be ensured for introducing shared static rqs that the LLDD cannot think that requests are associated with a specific HW queue. Signed-off-by: John Garry<redacted> --- block/blk-mq.c | 15 ++++++++++----- include/linux/blk-mq.h | 7 +++++++ 2 files changed, 17 insertions(+), 5 deletions(-)diff --git a/block/blk-mq.c b/block/blk-mq.c index f14cc2705f9b..4d6723cfa582 100644 --- a/block/blk-mq.c +++ b/block/blk-mq.c@@ -2427,13 +2427,15 @@ struct blk_mq_tags *blk_mq_alloc_rq_map(struct blk_mq_tag_set *set, static int blk_mq_init_request(struct blk_mq_tag_set *set, struct request *rq, unsigned int hctx_idx, int node) { - int ret; + int ret = 0; - if (set->ops->init_request) { + if (set->ops->init_request) ret = set->ops->init_request(set, rq, hctx_idx, node); - if (ret) - return ret; - } + else if (set->ops->init_request_no_hctx) + ret = set->ops->init_request_no_hctx(set, rq, node);
Hi Ming,
The only shared sbitmap user of SCSI does not use passed hctx_idx, not sure we need such new callback.
Sure, actually most versions of init_request callback don't use hctx_idx. Or numa_node arg.
If you really want to do this, just wondering why not pass '-1' as hctx_idx in case of shared sbitmap?
Yeah, I did consider that. hctx_idx is an unsigned, and I generally don't like -1U - but that's no big deal. But I also didn't like how it relies on the driver init_request callback to check the value, which changes the semantics. Obviously we don't add new versions of init_request for new block drivers which use shared sbitmap everyday, and any new ones would get it right. I suppose I can go the way you suggest - I just thought that this method was neat as well. Thanks, John