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btrfs: Don't hardcode the csum size in btrfs_ordered_sum_size 

Currently the function uses a hardcoded value for the checksum size of
a sector. This is fine, given that we currently support only a single
algorithm, whose checksum is 4 bytes == sizeof(u32). Despite not
having other algorithms, btrfs' design supports using a different
algorithm whith different space requirements. To future-proof the code
query the size of the currently used algorithm from the in-memory copy
of the super block. No functional changes.

Signed-off-by: Nikolay Borisov <nborisov@suse.com>
Reviewed-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: Su Yue <suy.fnst@cn.fujitsu.com>
Signed-off-by: David Sterba <dsterba@suse.com>
This commit is contained in:
Nikolay Borisov 2018-02-07 11:19:10 +02:00 committed by David Sterba
parent 97dc231e89
commit af89e0dc2c
1 changed files with 3 additions and 1 deletions
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4
fs/btrfs/ordered-data.h
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@ -151,7 +151,9 @@ static inline int btrfs_ordered_sum_size(struct btrfs_fs_info *fs_info,
unsigned long bytes) unsigned long bytes)
{ {
int num_sectors = (int)DIV_ROUND_UP(bytes, fs_info->sectorsize); int num_sectors = (int)DIV_ROUND_UP(bytes, fs_info->sectorsize);
return sizeof(struct btrfs_ordered_sum) + num_sectors * sizeof(u32); int csum_size = btrfs_super_csum_size(fs_info->super_copy);
return sizeof(struct btrfs_ordered_sum) + num_sectors * csum_size;
} }
static inline void static inline void