diff options
| -rw-r--r-- | fs/btrfs/inode.c | 57 |
1 files changed, 30 insertions, 27 deletions
diff --git a/fs/btrfs/inode.c b/fs/btrfs/inode.c index 97b601bec326..bb74a4181075 100644 --- a/fs/btrfs/inode.c +++ b/fs/btrfs/inode.c @@ -1542,30 +1542,17 @@ static void btrfs_split_extent_hook(struct inode *inode, u64 new_size; /* - * We need the largest size of the remaining extent to see if we - * need to add a new outstanding extent. Think of the following - * case - * - * [MEAX_EXTENT_SIZEx2 - 4k][4k] - * - * The new_size would just be 4k and we'd think we had enough - * outstanding extents for this if we only took one side of the - * split, same goes for the other direction. We need to see if - * the larger size still is the same amount of extents as the - * original size, because if it is we need to add a new - * outstanding extent. But if we split up and the larger size - * is less than the original then we are good to go since we've - * already accounted for the extra extent in our original - * accounting. + * See the explanation in btrfs_merge_extent_hook, the same + * applies here, just in reverse. */ new_size = orig->end - split + 1; - if ((split - orig->start) > new_size) - new_size = split - orig->start; - - num_extents = div64_u64(size + BTRFS_MAX_EXTENT_SIZE - 1, + num_extents = div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1, BTRFS_MAX_EXTENT_SIZE); - if (div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1, - BTRFS_MAX_EXTENT_SIZE) < num_extents) + new_size = split - orig->start; + num_extents += div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1, + BTRFS_MAX_EXTENT_SIZE); + if (div64_u64(size + BTRFS_MAX_EXTENT_SIZE - 1, + BTRFS_MAX_EXTENT_SIZE) >= num_extents) return; } @@ -1591,9 +1578,6 @@ static void btrfs_merge_extent_hook(struct inode *inode, if (!(other->state & EXTENT_DELALLOC)) return; - old_size = other->end - other->start + 1; - if (old_size < (new->end - new->start + 1)) - old_size = (new->end - new->start + 1); if (new->start > other->start) new_size = new->end - other->start + 1; else @@ -1608,13 +1592,32 @@ static void btrfs_merge_extent_hook(struct inode *inode, } /* - * If we grew by another max_extent, just return, we want to keep that - * reserved amount. + * We have to add up either side to figure out how many extents were + * accounted for before we merged into one big extent. If the number of + * extents we accounted for is <= the amount we need for the new range + * then we can return, otherwise drop. Think of it like this + * + * [ 4k][MAX_SIZE] + * + * So we've grown the extent by a MAX_SIZE extent, this would mean we + * need 2 outstanding extents, on one side we have 1 and the other side + * we have 1 so they are == and we can return. But in this case + * + * [MAX_SIZE+4k][MAX_SIZE+4k] + * + * Each range on their own accounts for 2 extents, but merged together + * they are only 3 extents worth of accounting, so we need to drop in + * this case. */ + old_size = other->end - other->start + 1; num_extents = div64_u64(old_size + BTRFS_MAX_EXTENT_SIZE - 1, BTRFS_MAX_EXTENT_SIZE); + old_size = new->end - new->start + 1; + num_extents += div64_u64(old_size + BTRFS_MAX_EXTENT_SIZE - 1, + BTRFS_MAX_EXTENT_SIZE); + if (div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1, - BTRFS_MAX_EXTENT_SIZE) > num_extents) + BTRFS_MAX_EXTENT_SIZE) >= num_extents) return; spin_lock(&BTRFS_I(inode)->lock); |