On Thu, Apr 07, 2022 at 09:25:40PM +0300, Andrew wrote:
>
> dd if=/dev/zero of=/tmp/ldiskfs bs=1M count=100
> mkfs.ext4 -O ea_inode /tmp/ldiskfs -J size=16 -I 512
>
> mkdir -p /tmp/ldiskfs_m
> mount -t ext4 /tmp/ldiskfs /tmp/ldiskfs_m -o loop,commit=600,no_mbcache
> touch /tmp/ldiskfs_m/file{1..1024}
>
> V=$(for i in `seq 60000`; do echo -n x ; done)
> V1="1$V"
> V2="2$V"
>
> for k in 1 2 3 4 5 6 7 8 9; do
> setfattr -n user.xattr -v $V /tmp/ldiskfs_m/file{1..1024}
> setfattr -n user.xattr -v $V1 /tmp/ldiskfs_m/file{1..1024} &
> setfattr -n user.xattr -v $V2 /tmp/ldiskfs_m/file{1024..1} &
> wait
> done
>
> umount /tmp/ldiskfs_m
Hi Andrew,
Thanks for the reproducer. I'll note that with the proposed patch, we
will allocate a *large* number of delayed_iput_work structure, and
most of the time, it is not necessary to do a delayed_iput, since we
won't actually be releasing the ea_inode. That only happens when the
ea_inode's refcount goes to zero.
So we could significantly reduce the overhead of this patch by
plumbing whether the refcount went to zero in
ext4_xattr_inode_update_ref() up through ext4_xattr_inode_dec_ref() to
ext4_xattr_set_entry(), and have it only call delayed_iput() if the
refcount went to zero.
Alternatively, we could add a function to fs/ext4/orphan.c which
checks whether the old_ea_inode is on the orphan list. And if it is
on the orphan list, then we know that the refcount must have gone to
zero, and so we need to call delayed_iput(). If the ea_inode isn't on
the orphan list, we can just use iput(), which will be fast, since it
will only be dropping i_count, and we don't need to worry about the
ea_inode getting truncate.
Could you take a look at this optimization, and then update the commit
description to explain what's happening, inlude the kernel stack, and
the reproducer?
Many thanks!
- Ted
