btrfs: keep track of the root owner for relocation reads

While testing the error paths in relocation, I hit the following lockdep
splat:

  ======================================================
  WARNING: possible circular locking dependency detected
  5.10.0-rc3+ #206 Not tainted
  ------------------------------------------------------
  btrfs-balance/1571 is trying to acquire lock:
  ffff8cdbcc8f77d0 (&head_ref->mutex){+.+.}-{3:3}, at: btrfs_lookup_extent_info+0x156/0x3b0

  but task is already holding lock:
  ffff8cdbc54adbf8 (btrfs-tree-00){++++}-{3:3}, at: __btrfs_tree_lock+0x27/0x100

  which lock already depends on the new lock.

  the existing dependency chain (in reverse order) is:

  -> #2 (btrfs-tree-00){++++}-{3:3}:
	 down_write_nested+0x43/0x80
	 __btrfs_tree_lock+0x27/0x100
	 btrfs_search_slot+0x248/0x890
	 relocate_tree_blocks+0x490/0x650
	 relocate_block_group+0x1ba/0x5d0
	 kretprobe_trampoline+0x0/0x50

  -> #1 (btrfs-csum-01){++++}-{3:3}:
	 down_read_nested+0x43/0x130
	 __btrfs_tree_read_lock+0x27/0x100
	 btrfs_read_lock_root_node+0x31/0x40
	 btrfs_search_slot+0x5ab/0x890
	 btrfs_del_csums+0x10b/0x3c0
	 __btrfs_free_extent+0x49d/0x8e0
	 __btrfs_run_delayed_refs+0x283/0x11f0
	 btrfs_run_delayed_refs+0x86/0x220
	 btrfs_start_dirty_block_groups+0x2ba/0x520
	 kretprobe_trampoline+0x0/0x50

  -> #0 (&head_ref->mutex){+.+.}-{3:3}:
	 __lock_acquire+0x1167/0x2150
	 lock_acquire+0x116/0x3e0
	 __mutex_lock+0x7e/0x7b0
	 btrfs_lookup_extent_info+0x156/0x3b0
	 walk_down_proc+0x1c3/0x280
	 walk_down_tree+0x64/0xe0
	 btrfs_drop_subtree+0x182/0x260
	 do_relocation+0x52e/0x660
	 relocate_tree_blocks+0x2ae/0x650
	 relocate_block_group+0x1ba/0x5d0
	 kretprobe_trampoline+0x0/0x50

  other info that might help us debug this:

  Chain exists of:
    &head_ref->mutex --> btrfs-csum-01 --> btrfs-tree-00

   Possible unsafe locking scenario:

	 CPU0                    CPU1
	 ----                    ----
    lock(btrfs-tree-00);
				 lock(btrfs-csum-01);
				 lock(btrfs-tree-00);
    lock(&head_ref->mutex);

   *** DEADLOCK ***

  5 locks held by btrfs-balance/1571:
   #0: ffff8cdb89749ff8 (&fs_info->delete_unused_bgs_mutex){+.+.}-{3:3}, at: btrfs_balance+0x563/0xf40
   #1: ffff8cdb89748838 (&fs_info->cleaner_mutex){+.+.}-{3:3}, at: btrfs_relocate_block_group+0x156/0x300
   #2: ffff8cdbc2c16650 (sb_internal#2){.+.+}-{0:0}, at: start_transaction+0x413/0x5c0
   #3: ffff8cdbc135f538 (btrfs-treloc-01){+.+.}-{3:3}, at: __btrfs_tree_lock+0x27/0x100
   #4: ffff8cdbc54adbf8 (btrfs-tree-00){++++}-{3:3}, at: __btrfs_tree_lock+0x27/0x100

  stack backtrace:
  CPU: 1 PID: 1571 Comm: btrfs-balance Not tainted 5.10.0-rc3+ #206
  Hardware name: QEMU Standard PC (Q35 + ICH9, 2009), BIOS 1.13.0-2.fc32 04/01/2014
  Call Trace:
   dump_stack+0x8b/0xb0
   check_noncircular+0xcf/0xf0
   ? trace_call_bpf+0x139/0x260
   __lock_acquire+0x1167/0x2150
   lock_acquire+0x116/0x3e0
   ? btrfs_lookup_extent_info+0x156/0x3b0
   __mutex_lock+0x7e/0x7b0
   ? btrfs_lookup_extent_info+0x156/0x3b0
   ? btrfs_lookup_extent_info+0x156/0x3b0
   ? release_extent_buffer+0x124/0x170
   ? _raw_spin_unlock+0x1f/0x30
   ? release_extent_buffer+0x124/0x170
   btrfs_lookup_extent_info+0x156/0x3b0
   walk_down_proc+0x1c3/0x280
   walk_down_tree+0x64/0xe0
   btrfs_drop_subtree+0x182/0x260
   do_relocation+0x52e/0x660
   relocate_tree_blocks+0x2ae/0x650
   ? add_tree_block+0x149/0x1b0
   relocate_block_group+0x1ba/0x5d0
   elfcorehdr_read+0x40/0x40
   ? elfcorehdr_read+0x40/0x40
   ? btrfs_balance+0x796/0xf40
   ? __kthread_parkme+0x66/0x90
   ? btrfs_balance+0xf40/0xf40
   ? balance_kthread+0x37/0x50
   ? kthread+0x137/0x150
   ? __kthread_bind_mask+0x60/0x60
   ? ret_from_fork+0x1f/0x30

As you can see this is bogus, we never take another tree's lock under
the csum lock.  This happens because sometimes we have to read tree
blocks from disk without knowing which root they belong to during
relocation.  We defaulted to an owner of 0, which translates to an fs
tree.  This is fine as all fs trees have the same class, but obviously
isn't fine if the block belongs to a COW only tree.

Thankfully COW only trees only have their owners root as a reference to
them, and since we already look up the extent information during
relocation, go ahead and check and see if this block might belong to a
COW only tree, and if so save the owner in the tree_block struct.  This
allows us to read_tree_block with the proper owner, which gets rid of
this lockdep splat.

Signed-off-by: Josef Bacik <josef@toxicpanda.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>

1 files changed, 43 insertions, 3 deletions

diff --git a/fs/btrfs/relocation.c b/fs/btrfs/relocation.c
index 8e51b39cbfbb..9f2289bcdde6 100644
--- a/fs/btrfs/relocation.c
+++ b/fs/btrfs/relocation.c
@@ -97,6 +97,7 @@ struct tree_block {
 		struct rb_node rb_node;
 		u64 bytenr;
 	}; /* Use rb_simple_node for search/insert */
+	u64 owner;
 	struct btrfs_key key;
 	unsigned int level:8;
 	unsigned int key_ready:1;
@@ -2393,8 +2394,8 @@ static int get_tree_block_key(struct btrfs_fs_info *fs_info,
 {
 	struct extent_buffer *eb;
 
-	eb = read_tree_block(fs_info, block->bytenr, 0, block->key.offset,
-			     block->level, NULL);
+	eb = read_tree_block(fs_info, block->bytenr, block->owner,
+			     block->key.offset, block->level, NULL);
 	if (IS_ERR(eb)) {
 		return PTR_ERR(eb);
 	} else if (!extent_buffer_uptodate(eb)) {
@@ -2493,7 +2494,8 @@ int relocate_tree_blocks(struct btrfs_trans_handle *trans,
 	/* Kick in readahead for tree blocks with missing keys */
 	rbtree_postorder_for_each_entry_safe(block, next, blocks, rb_node) {
 		if (!block->key_ready)
-			btrfs_readahead_tree_block(fs_info, block->bytenr, 0, 0,
+			btrfs_readahead_tree_block(fs_info, block->bytenr,
+						   block->owner, 0,
 						   block->level);
 	}
 
@@ -2801,21 +2803,58 @@ static int add_tree_block(struct reloc_control *rc,
 	u32 item_size;
 	int level = -1;
 	u64 generation;
+	u64 owner = 0;
 
 	eb =  path->nodes[0];
 	item_size = btrfs_item_size_nr(eb, path->slots[0]);
 
 	if (extent_key->type == BTRFS_METADATA_ITEM_KEY ||
 	    item_size >= sizeof(*ei) + sizeof(*bi)) {
+		unsigned long ptr = 0, end;
+
 		ei = btrfs_item_ptr(eb, path->slots[0],
 				struct btrfs_extent_item);
+		end = (unsigned long)ei + item_size;
 		if (extent_key->type == BTRFS_EXTENT_ITEM_KEY) {
 			bi = (struct btrfs_tree_block_info *)(ei + 1);
 			level = btrfs_tree_block_level(eb, bi);
+			ptr = (unsigned long)(bi + 1);
 		} else {
 			level = (int)extent_key->offset;
+			ptr = (unsigned long)(ei + 1);
 		}
 		generation = btrfs_extent_generation(eb, ei);
+
+		/*
+		 * We're reading random blocks without knowing their owner ahead
+		 * of time.  This is ok most of the time, as all reloc roots and
+		 * fs roots have the same lock type.  However normal trees do
+		 * not, and the only way to know ahead of time is to read the
+		 * inline ref offset.  We know it's an fs root if
+		 *
+		 * 1. There's more than one ref.
+		 * 2. There's a SHARED_DATA_REF_KEY set.
+		 * 3. FULL_BACKREF is set on the flags.
+		 *
+		 * Otherwise it's safe to assume that the ref offset == the
+		 * owner of this block, so we can use that when calling
+		 * read_tree_block.
+		 */
+		if (btrfs_extent_refs(eb, ei) == 1 &&
+		    !(btrfs_extent_flags(eb, ei) &
+		      BTRFS_BLOCK_FLAG_FULL_BACKREF) &&
+		    ptr < end) {
+			struct btrfs_extent_inline_ref *iref;
+			int type;
+
+			iref = (struct btrfs_extent_inline_ref *)ptr;
+			type = btrfs_get_extent_inline_ref_type(eb, iref,
+							BTRFS_REF_TYPE_BLOCK);
+			if (type == BTRFS_REF_TYPE_INVALID)
+				return -EINVAL;
+			if (type == BTRFS_TREE_BLOCK_REF_KEY)
+				owner = btrfs_extent_inline_ref_offset(eb, iref);
+		}
 	} else if (unlikely(item_size == sizeof(struct btrfs_extent_item_v0))) {
 		btrfs_print_v0_err(eb->fs_info);
 		btrfs_handle_fs_error(eb->fs_info, -EINVAL, NULL);
@@ -2837,6 +2876,7 @@ static int add_tree_block(struct reloc_control *rc,
 	block->key.offset = generation;
 	block->level = level;
 	block->key_ready = 0;
+	block->owner = owner;
 
 	rb_node = rb_simple_insert(blocks, block->bytenr, &block->rb_node);
 	if (rb_node)