btrfs: skip hole detection during full fsync for files without holes

If we the no-holes feature is enabled (a default since btrfs-progs 5.15),
when doing a full fsync we always iterate of all leaves in the subvolume
root that contain file extent items in order to detect holes between them.

This can take a lot of time for files with a large number of extents.
But if we know there are no prealloc extents and the amount of space
(uncompressed space) is greater than or equals to the i_size of the
inode, then we cannot have holes and therefore avoid searching for
them. So skip the search if those conditions are met.

The following test script was used:

  $ cat test.sh
  #!/bin/bash

  MNT=/mnt/nullb0
  DEV=/dev/nullb0

  umount $MNT &> /dev/null
  mkfs.btrfs -f $DEV
  mount $DEV $MNT

  # 256M gives 64K extents of 4K each.
  FILE_SIZE=$((256 * 1024 * 1024))
  touch $MNT/foobar

  for ((i = 0; i < $FILE_SIZE; i += 8192)); do
      xfs_io -c "pwrite -S 0xab $i 4K" $MNT/foobar > /dev/null
  done

  xfs_io -c "fsync" $MNT/foobar

  for ((i = 4096; i < $FILE_SIZE; i += 8192)); do
     xfs_io -c "pwrite -S 0xab $i 4K" $MNT/foobar > /dev/null
  done

  # unmount and mount, clear caches and ensure the next fsync is a
  # full sync.
  umount $MNT
  mount $DEV $MNT

  # Do some change to the file in order to fsync.
  xfs_io -c "pwrite -S 0xcd 0 4K" $MNT/foobar > /dev/null

  T0=$(date +%s%N)
  xfs_io -c "fsync" $MNT/foobar
  T1=$(date +%s%N)

  echo
  echo "Took $(( (T1 - T0) / 1000 ))us"

  umount $MNT

Before this change:

  Took 28721us

After this change:

  Took 5453us

That's about 5.3x times faster.

Reviewed-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>

0 files changed, 0 insertions, 0 deletions