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>
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@@ -416,7 +416,7 @@ int convert_free_space_to_extents(struct btrfs_trans_handle *trans,
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btrfs_mark_buffer_dirty(leaf);
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btrfs_release_path(path);
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nrbits = div_u64(block_group->length, block_group->fs_info->sectorsize);
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nrbits = block_group->length >> block_group->fs_info->sectorsize_bits;
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start_bit = find_next_bit_le(bitmap, nrbits, 0);
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while (start_bit < nrbits) {
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@@ -540,8 +540,8 @@ static void free_space_set_bits(struct btrfs_block_group *block_group,
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end = found_end;
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ptr = btrfs_item_ptr_offset(leaf, path->slots[0]);
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first = div_u64(*start - found_start, fs_info->sectorsize);
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last = div_u64(end - found_start, fs_info->sectorsize);
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first = (*start - found_start) >> fs_info->sectorsize_bits;
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last = (end - found_start) >> fs_info->sectorsize_bits;
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if (bit)
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extent_buffer_bitmap_set(leaf, ptr, first, last - first);
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else
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