Re: [PATCH] btrfs: extract the parity scrub code into a helper

[David Sterba <dsterba@suse.cz>]

On Thu, Nov 06, 2025 at 01:32:05PM +1030, Qu Wenruo wrote:
> The function scrub_raid56_parity_stripe() is handling the partity stripe
> by the following steps:
> 
> - Scrub each data stripes
>   And make sure everything is fine in each data stripe
> 
> - Cache the data stripe into the raid bio
> 
> - Use the cached raid bio to scrub the target parity stripe
> 
> Extract the last two steps into a new helper,
> scrub_radi56_cached_parity(), as a cleanup and make the error handling
> more straightforward.
> 
> With the following minor cleanups:
> 
> - Use on-stack bio structure
>   The bio is always empty thus we do not need any bio vector nor the
>   block device. Thus there is no need to allocate a bio, the on-stack
>   one is more than enough to cut it.
> 
> - Remove the unnecessary btrfs_put_bioc() call if btrfs_map_block()
>   failed
>   If btrfs_map_block() is failed, @bioc_ret will not be touched thus
>   there is no need to call btrfs_put_bioc() in this case.
> 
> - Use a proper out: tag to do the cleanup
>   Now the error cleanup is much shorter and simpler, just
>   btrfs_bio_counter_dec() and bio_uninit().
> 
> Signed-off-by: Qu Wenruo <wqu@suse.com>
> ---
>  fs/btrfs/scrub.c | 90 ++++++++++++++++++++++++++++--------------------
>  1 file changed, 52 insertions(+), 38 deletions(-)
> 
> diff --git a/fs/btrfs/scrub.c b/fs/btrfs/scrub.c
> index e3612202ba55..8c360d941bd5 100644
> --- a/fs/btrfs/scrub.c
> +++ b/fs/btrfs/scrub.c
> @@ -2113,6 +2113,56 @@ static int should_cancel_scrub(const struct scrub_ctx *sctx)
>  	return 0;
>  }
>  
> +static int scrub_raid56_cached_parity(struct scrub_ctx *sctx,
> +				      struct btrfs_device *scrub_dev,
> +				      struct btrfs_chunk_map *map,
> +				      u64 full_stripe_start,
> +				      unsigned long *extent_bitmap)
> +{
> +	DECLARE_COMPLETION_ONSTACK(io_done);
> +	struct btrfs_fs_info *fs_info = sctx->fs_info;
> +	struct btrfs_io_context *bioc = NULL;
> +	struct btrfs_raid_bio *rbio;
> +	struct bio bio;
> +	const int data_stripes = nr_data_stripes(map);
> +	u64 length = btrfs_stripe_nr_to_offset(data_stripes);
> +	int ret;
> +
> +	bio_init(&bio, NULL, NULL, 0, REQ_OP_READ);
> +	bio.bi_iter.bi_sector = full_stripe_start >> SECTOR_SHIFT;
> +	bio.bi_private = &io_done;
> +	bio.bi_end_io = raid56_scrub_wait_endio;
> +
> +	btrfs_bio_counter_inc_blocked(fs_info);
> +	ret = btrfs_map_block(fs_info, BTRFS_MAP_WRITE, full_stripe_start,
> +			      &length, &bioc, NULL, NULL);
> +	if (ret < 0)
> +		goto out;
> +	/* For RAID56 write there must be an @bioc allocated. */
> +	ASSERT(bioc);
> +	rbio = raid56_parity_alloc_scrub_rbio(&bio, bioc, scrub_dev, extent_bitmap,
> +				BTRFS_STRIPE_LEN >> fs_info->sectorsize_bits);
> +	btrfs_put_bioc(bioc);
> +	if (!rbio) {
> +		ret = -ENOMEM;
> +		goto out;
> +	}
> +	/* Use the recovered stripes as cache to avoid read them from disk again. */
> +	for (int i = 0; i < data_stripes; i++) {
> +		struct scrub_stripe *stripe = &sctx->raid56_data_stripes[i];
> +
> +		raid56_parity_cache_data_folios(rbio, stripe->folios,
> +				full_stripe_start + (i << BTRFS_STRIPE_LEN_SHIFT));
> +	}
> +	raid56_parity_submit_scrub_rbio(rbio);
> +	wait_for_completion_io(&io_done);
> +	ret = blk_status_to_errno(bio.bi_status);
> +out:
> +	btrfs_bio_counter_dec(fs_info);
> +	bio_uninit(&bio);
> +	return ret;
> +}
> +
>  static int scrub_raid56_parity_stripe(struct scrub_ctx *sctx,
>  				      struct btrfs_device *scrub_dev,
>  				      struct btrfs_block_group *bg,
> @@ -2121,16 +2171,12 @@ static int scrub_raid56_parity_stripe(struct scrub_ctx *sctx,
>  {
>  	DECLARE_COMPLETION_ONSTACK(io_done);

This should be deleted as well, as it's in scrub_raid56_cached_parity()

The stack meter says that the new function adds 240 bytes (and has
dynamic stack size) while scrub_raid56_parity_stripe() shrinks only by
24 bytes. So this basically adds 240 - 24 = 216 bytes to the stack.

With the completion removed is another -32 bytes it's down to 184. The
on-stack bio is 112 bytes from that, 184 - 112 = 72 for remaining
variables.