Re: [PATCH 0/3] btrfs: fixes for relocation to avoid KASAN reports

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On 2020/1/4 上午12:15, David Sterba wrote:
> On Fri, Jan 03, 2020 at 04:52:59PM +0100, David Sterba wrote:
>> So it's one bit vs refcount and a lock. For the backports I'd go with
>> the bit, but this needs the barriers as mentioned in my previous reply.
>> Can you please update the patches?
> 
> The idea is in the diff below (compile tested only). I found one more
> case that was not addressed by your patches, it's in
> btrfs_update_reloc_root.
> 
> Given that the type of the fix is the same, I'd rather do that in one
> patch. The reported stack traces are more or less the same.
> 
> diff --git a/fs/btrfs/relocation.c b/fs/btrfs/relocation.c
> index af4dd49a71c7..aeba3a7506e1 100644
> --- a/fs/btrfs/relocation.c
> +++ b/fs/btrfs/relocation.c
> @@ -517,6 +517,15 @@ static int update_backref_cache(struct btrfs_trans_handle *trans,
>  	return 1;
>  }
>  
> +static bool have_reloc_root(struct btrfs_root *root)
> +{
> +	smp_mb__before_atomic();

Mind to explain why the before_atomic() is needed?

Is it just paired with smp_mb__after_atomic() for the
set_bit()/clear_bit() part?

>  	reloc_root = root->reloc_root;
> @@ -1489,6 +1498,7 @@ int btrfs_update_reloc_root(struct btrfs_trans_handle *trans,
>  	if (fs_info->reloc_ctl->merge_reloc_tree &&
>  	    btrfs_root_refs(root_item) == 0) {
>  		set_bit(BTRFS_ROOT_DEAD_RELOC_TREE, &root->state);
> +		smp_mb__after_atomic();

I get the point here, to make sure all other users see this bit change.

>  		__del_reloc_root(reloc_root);

Interestingly in that function we immediately triggers spin_lock() which
implies memory barrier.
(Not an excuse to skip memory barrier anyway)

>  	}
>  
> @@ -2201,6 +2211,7 @@ static int clean_dirty_subvols(struct reloc_control *rc)
>  				if (ret2 < 0 && !ret)
>  					ret = ret2;
>  			}
> +			smp_mb__before_atomic();
>  			clear_bit(BTRFS_ROOT_DEAD_RELOC_TREE, &root->state);

I guess this should be a smp_mb__after_atomic();

>  			btrfs_put_fs_root(root);

And btrfs_put_fs_root() triggers a release memory ordering.

So it looks memory order is not completely screwed up before, completely
by pure luck...

Thanks,
Qu

>  		} else {
> @@ -4730,7 +4741,7 @@ void btrfs_reloc_pre_snapshot(struct btrfs_pending_snapshot *pending,
>  	struct btrfs_root *root = pending->root;
>  	struct reloc_control *rc = root->fs_info->reloc_ctl;
>  
> -	if (!root->reloc_root || !rc)
> +	if (!rc || !have_reloc_root(root))
>  		return;
>  
>  	if (!rc->merge_reloc_tree)
> @@ -4764,7 +4775,7 @@ int btrfs_reloc_post_snapshot(struct btrfs_trans_handle *trans,
>  	struct reloc_control *rc = root->fs_info->reloc_ctl;
>  	int ret;
>  
> -	if (!root->reloc_root || !rc)
> +	if (!rc || !have_reloc_root(root))
>  		return 0;
>  
>  	rc = root->fs_info->reloc_ctl;
> 

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