btrfs: use precalculated sectorsize_bits from fs_info
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>
diff --git a/fs/btrfs/disk-io.c b/fs/btrfs/disk-io.c
index ad526aa..25889a2 100644
--- a/fs/btrfs/disk-io.c
+++ b/fs/btrfs/disk-io.c
@@ -2814,6 +2814,7 @@ void btrfs_init_fs_info(struct btrfs_fs_info *fs_info)
/* Usable values until the real ones are cached from the superblock */
fs_info->nodesize = 4096;
fs_info->sectorsize = 4096;
+ fs_info->sectorsize_bits = ilog2(4096);
fs_info->stripesize = 4096;
spin_lock_init(&fs_info->swapfile_pins_lock);
@@ -3078,6 +3079,7 @@ int __cold open_ctree(struct super_block *sb, struct btrfs_fs_devices *fs_device
/* Cache block sizes */
fs_info->nodesize = nodesize;
fs_info->sectorsize = sectorsize;
+ fs_info->sectorsize_bits = ilog2(sectorsize);
fs_info->stripesize = stripesize;
/*
