Add a min size parameter to btrfs_alloc_extent

On huge machines, delayed allocation may try to allocate massive extents.
This change allows btrfs_alloc_extent to return something smaller than
the caller asked for, and the data allocation routines will loop over
the allocations until it fills the whole delayed alloc.

Signed-off-by: Chris Mason <chris.mason@oracle.com>

1 files changed, 2 insertions, 0 deletions

diff --git a/fs/btrfs/inode.c b/fs/btrfs/inode.c
index 46f54dfa2b0a..7869c0157a1f 100644
--- a/fs/btrfs/inode.c
+++ b/fs/btrfs/inode.c
@@ -125,6 +125,7 @@ static int cow_file_range(struct inode *inode, u64 start, u64 end)
         while(num_bytes > 0) {
                 cur_alloc_size = min(num_bytes, root->fs_info->max_extent);
                 ret = btrfs_alloc_extent(trans, root, cur_alloc_size,
+                                         root->sectorsize,
                                          root->root_key.objectid,
                                          trans->transid,
                                          inode->i_ino, start, 0,
@@ -133,6 +134,7 @@ static int cow_file_range(struct inode *inode, u64 start, u64 end)
                         WARN_ON(1);
                         goto out;
                 }
+                cur_alloc_size = ins.offset;
                 ret = btrfs_insert_file_extent(trans, root, inode->i_ino,
                                                start, ins.objectid, ins.offset,
                                                ins.offset);