@@ -9599,6 +9599,7 @@ static int __btrfs_prealloc_file_range(struct inode *inode, int mode,
u64 cur_offset = start;
u64 i_size;
u64 cur_bytes;
+ u64 last_alloc = (u64)-1;
int ret = 0;
bool own_trans = true;
@@ -9615,6 +9616,13 @@ static int __btrfs_prealloc_file_range(struct inode *inode, int mode,
cur_bytes = min(num_bytes, 256ULL * 1024 * 1024);
cur_bytes = max(cur_bytes, min_size);
+ /*
+ * If we are severely fragmented we could end up with really
+ * small allocations, so if the allocator is returning small
+ * chunks lets make its job easier by only searching for those
+ * sized chunks.
+ */
+ cur_bytes = min(cur_bytes, last_alloc);
ret = btrfs_reserve_extent(root, cur_bytes, min_size, 0,
*alloc_hint, &ins, 1, 0);
if (ret) {
@@ -9623,6 +9631,7 @@ static int __btrfs_prealloc_file_range(struct inode *inode, int mode,
break;
}
+ last_alloc = ins.offset;
ret = insert_reserved_file_extent(trans, inode,
cur_offset, ins.objectid,
ins.offset, ins.offset,
If we are heavily fragmented we will continually try to prealloc the largest extent size we can every time we call btrfs_reserve_extent. This can be very expensive when we are heavily fragmented, burning lots of CPU cycles and loops through the allocator. So instead notice when we get a smaller chunk from the allocator than what we specified and use this as the new maximum size we try to allocate. Thanks, Signed-off-by: Josef Bacik <jbacik@fb.com> --- fs/btrfs/inode.c | 9 +++++++++ 1 file changed, 9 insertions(+)