Re: [PATCH 08/63] xfs: account for the refcount btree in the alloc/free log reservation

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On Tue, Sep 27, 2016 at 07:54:20PM -0700, Darrick J. Wong wrote:
> Every time we allocate or free an extent, we might need to split the
> refcount btree.  Reserve some blocks in the transaction to handle
> this possibility.
> 

I'm probably not far along enough to grok how this all works, but I'm a
bit confused by the requirement here. Why is generic block
allocation/free a refcountbt matter? E.g., the function below is used
for things like inode chunk allocation and whatnot, aren't those
irrelevant to the refcountbt?

Brian

> (Reproduced by generic/167 over NFS atop XFS)
> 
> Signed-off-by: Christoph Hellwig <hch@xxxxxx>
> [darrick.wong@xxxxxxxxxx: add commit message]
> Signed-off-by: Darrick J. Wong <darrick.wong@xxxxxxxxxx>
> ---
>  fs/xfs/libxfs/xfs_trans_resv.c |    5 ++++-
>  1 file changed, 4 insertions(+), 1 deletion(-)
> 
> 
> diff --git a/fs/xfs/libxfs/xfs_trans_resv.c b/fs/xfs/libxfs/xfs_trans_resv.c
> index 7c840e1..a59838f 100644
> --- a/fs/xfs/libxfs/xfs_trans_resv.c
> +++ b/fs/xfs/libxfs/xfs_trans_resv.c
> @@ -67,7 +67,8 @@ xfs_calc_buf_res(
>   * Per-extent log reservation for the btree changes involved in freeing or
>   * allocating an extent.  In classic XFS there were two trees that will be
>   * modified (bnobt + cntbt).  With rmap enabled, there are three trees
> - * (rmapbt).  The number of blocks reserved is based on the formula:
> + * (rmapbt).  With reflink, there are four trees (refcountbt).  The number of
> + * blocks reserved is based on the formula:
>   *
>   * num trees * ((2 blocks/level * max depth) - 1)
>   *
> @@ -83,6 +84,8 @@ xfs_allocfree_log_count(
>  	blocks = num_ops * 2 * (2 * mp->m_ag_maxlevels - 1);
>  	if (xfs_sb_version_hasrmapbt(&mp->m_sb))
>  		blocks += num_ops * (2 * mp->m_rmap_maxlevels - 1);
> +	if (xfs_sb_version_hasreflink(&mp->m_sb))
> +		blocks += num_ops * (2 * mp->m_refc_maxlevels - 1);
>  
>  	return blocks;
>  }
> 
> --
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