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btrfs: avoid extra memory allocation when copying free space cache
At copy_free_space_cache(), we add a new entry to the block group's ctl before we free the entry from the temporary ctl. Adding a new entry requires the allocation of a new struct btrfs_free_space, so we can avoid a temporary extra allocation by freeing the entry from the temporary ctl before we add a new entry to the main ctl, which possibly also reduces the chances for a memory allocation failure in case of very high memory pressure. So just do that. Reviewed-by: Anand Jain <anand.jain@oracle.com> Signed-off-by: Filipe Manana <fdmanana@suse.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
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1 changed files with 4 additions and 2 deletions
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@ -923,10 +923,12 @@ static int copy_free_space_cache(struct btrfs_block_group *block_group,
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while (!ret && (n = rb_first(&ctl->free_space_offset)) != NULL) {
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info = rb_entry(n, struct btrfs_free_space, offset_index);
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if (!info->bitmap) {
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const u64 offset = info->offset;
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const u64 bytes = info->bytes;
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unlink_free_space(ctl, info, true);
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ret = btrfs_add_free_space(block_group, info->offset,
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info->bytes);
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kmem_cache_free(btrfs_free_space_cachep, info);
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ret = btrfs_add_free_space(block_group, offset, bytes);
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} else {
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u64 offset = info->offset;
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u64 bytes = ctl->unit;
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