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Add a min size parameter to btrfs_alloc_extent

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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>
This commit is contained in:
Chris Mason
2008-04-14 09:46:10 -04:00
parent 73f61b2a64
commit 98d20f67cf
3 changed files with 13 additions and 4 deletions
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2
fs/btrfs/inode.c
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@@ -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);