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btrfs: use precalculated sectorsize_bits from fs_info

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We do a lot of calculations where we divide or multiply by sectorsize.
We also know and make sure that sectorsize is a power of two, so this
means all divisions can be turned to shifts and avoid eg. expensive
u64/u32 divisions.

The type is u32 as it's more register friendly on x86_64 compared to u8
and the resulting assembly is smaller (movzbl vs movl).

There's also superblock s_blocksize_bits but it's usually one more
pointer dereference farther than fs_info.

Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Signed-off-by: David Sterba <dsterba@suse.com>
This commit is contained in:
David Sterba
2020-07-01 20:45:04 +02:00
parent e940e9a7c7
commit ab108d992b
8 changed files with 17 additions and 13 deletions
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3
fs/btrfs/ordered-data.c
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@@ -868,7 +868,6 @@ int btrfs_find_ordered_sum(struct btrfs_inode *inode, u64 offset,
struct btrfs_ordered_inode_tree *tree = &inode->ordered_tree;
unsigned long num_sectors;
unsigned long i;
u32 sectorsize = btrfs_inode_sectorsize(inode);
const u8 blocksize_bits = inode->vfs_inode.i_sb->s_blocksize_bits;
const u16 csum_size = btrfs_super_csum_size(fs_info->super_copy);
int index = 0;
@@ -890,7 +889,7 @@ int btrfs_find_ordered_sum(struct btrfs_inode *inode, u64 offset,
index += (int)num_sectors * csum_size;
if (index == len)
goto out;
disk_bytenr += num_sectors * sectorsize;
disk_bytenr += num_sectors * fs_info->sectorsize;
}
}
out: