linux-kernel - Re: [PATCH v2 4/8] block: introduce bio_required_sector

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Date:   Thu, 15 Apr 2021 12:44:30 -0700
From:   Eric Biggers <ebiggers@...nel.org>
To:     Satya Tangirala <satyat@...gle.com>
Cc:     linux-block@...r.kernel.org, linux-kernel@...r.kernel.org,
        Jens Axboe <axboe@...nel.dk>
Subject: Re: [PATCH v2 4/8] block: introduce bio_required_sector_alignment()

On Thu, Mar 25, 2021 at 09:26:05PM +0000, Satya Tangirala wrote:
> This function returns the required alignment for the number of sectors in
> a bio. In particular, the number of sectors passed to bio_split() must be
> aligned to this value.
> 
> Signed-off-by: Satya Tangirala <satyat@...gle.com>
> ---
>  block/blk.h | 12 ++++++++++++
>  1 file changed, 12 insertions(+)
> 
> diff --git a/block/blk.h b/block/blk.h
> index 3b53e44b967e..5ef207a6d34c 100644
> --- a/block/blk.h
> +++ b/block/blk.h
> @@ -261,6 +261,18 @@ static inline unsigned int bio_allowed_max_sectors(struct request_queue *q)
>  	return round_down(UINT_MAX, queue_logical_block_size(q)) >> 9;
>  }
>  
> +/*
> + * The required sector alignment for a bio. The number of sectors in any bio
> + * that's constructed/split must be aligned to this value.

What does "constructed" mean in this context?

> + */
> +static inline unsigned int bio_required_sector_alignment(struct bio *bio)
> +{
> +	struct request_queue *q = bio->bi_bdev->bd_disk->queue;
> +
> +	return max(queue_logical_block_size(q) >> SECTOR_SHIFT,
> +		   blk_crypto_bio_sectors_alignment(bio));
> +}

It would be slightly simpler to use bdev_logical_block_size(bio->bi_bdev)
instead of queue_logical_block_size(bio->bi_bdev->bd_disk->queue).

Also, if there's interest in avoiding the branch implied by the max() here, it
would be possible to take advantage of the values being powers of 2 as follows:

static inline unsigned int bio_required_sector_alignment(struct bio *bio)
{
	unsigned int alignmask =
		(bdev_logical_block_size(q) >> SECTOR_SHIFT) - 1;

	alignmask |= blk_crypto_bio_sectors_alignment(bio) - 1;

	return alignmask + 1;
}