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Btrfs: adjust outstanding_extents counter properly when dio write is split

Currently how btrfs dio deals with split dio write is not good
enough if dio write is split into several segments due to the
lack of contiguous space, a large dio write like 'dd bs=1G count=1'
can end up with incorrect outstanding_extents counter and endio
would complain loudly with an assertion.

This fixes the problem by compensating the outstanding_extents
counter in inode if a large dio write gets split.

Reported-by: Anand Jain <anand.jain@oracle.com>
Tested-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Liu Bo <bo.li.liu@oracle.com>
Signed-off-by: David Sterba <dsterba@suse.com>
Liu Bo 8 years ago
parent
commit
c2931667c8
1 changed files with 9 additions and 2 deletions
  1. 9 2
      fs/btrfs/inode.c

+ 9 - 2
fs/btrfs/inode.c

@@ -7623,11 +7623,18 @@ static void adjust_dio_outstanding_extents(struct inode *inode,
 	 * within our reservation, otherwise we need to adjust our inode
 	 * within our reservation, otherwise we need to adjust our inode
 	 * counter appropriately.
 	 * counter appropriately.
 	 */
 	 */
-	if (dio_data->outstanding_extents) {
+	if (dio_data->outstanding_extents >= num_extents) {
 		dio_data->outstanding_extents -= num_extents;
 		dio_data->outstanding_extents -= num_extents;
 	} else {
 	} else {
+		/*
+		 * If dio write length has been split due to no large enough
+		 * contiguous space, we need to compensate our inode counter
+		 * appropriately.
+		 */
+		u64 num_needed = num_extents - dio_data->outstanding_extents;
+
 		spin_lock(&BTRFS_I(inode)->lock);
 		spin_lock(&BTRFS_I(inode)->lock);
-		BTRFS_I(inode)->outstanding_extents += num_extents;
+		BTRFS_I(inode)->outstanding_extents += num_needed;
 		spin_unlock(&BTRFS_I(inode)->lock);
 		spin_unlock(&BTRFS_I(inode)->lock);
 	}
 	}
 }
 }