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btrfs: remove SCRUB_MAX_SECTORS_PER_BLOCK

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The last user of this macro is removed in commit 001e3fc263 ("btrfs:
scrub: remove scrub_block and scrub_sector structures").

Now that macro is only utilized in an ASSERT(), which no longer makes
much sense.

Just remove it completely.

Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
This commit is contained in:
Qu Wenruo 2026-07-05 21:24:46 +09:30 committed by David Sterba
parent a503aa0610
commit ecfe11e8f4
1 changed files with 0 additions and 14 deletions
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14
fs/btrfs/scrub.c
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@ -57,12 +57,6 @@ struct scrub_ctx;
#define SCRUB_TOTAL_STRIPES (SCRUB_GROUPS_PER_SCTX * SCRUB_STRIPES_PER_GROUP)
/*
* The following value times PAGE_SIZE needs to be large enough to match the
* largest node/leaf/sector size that shall be supported.
*/
#define SCRUB_MAX_SECTORS_PER_BLOCK (BTRFS_MAX_METADATA_BLOCKSIZE / SZ_4K)
/* Represent one sector and its needed info to verify the content. */
struct scrub_sector_verification {
union {
@ -3091,14 +3085,6 @@ int btrfs_scrub_dev(struct btrfs_fs_info *fs_info, u64 devid, u64 start,
/* At mount time we have ensured nodesize is in the range of [4K, 64K]. */
ASSERT(fs_info->nodesize <= BTRFS_STRIPE_LEN);
/*
* SCRUB_MAX_SECTORS_PER_BLOCK is calculated using the largest possible
* value (max nodesize / min sectorsize), thus nodesize should always
* be fine.
*/
ASSERT(fs_info->nodesize <=
SCRUB_MAX_SECTORS_PER_BLOCK << fs_info->sectorsize_bits);
/* Allocate outside of device_list_mutex */
sctx = scrub_setup_ctx(fs_info, is_dev_replace);
if (IS_ERR(sctx))