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| author | Tao Ma <tao.ma@oracle.com> | 2010-07-22 13:56:45 +0800 |
|---|---|---|
| committer | Joel Becker <joel.becker@oracle.com> | 2010-08-07 10:44:49 -0700 |
| commit | 8a2e70c40ff58f82dde67770e6623ca45f0cb0c8 (patch) | |
| tree | ce05111dbb00fae81d0c0b7726ddac5c0117ec34 /fs/ocfs2/refcounttree.c | |
| parent | 7beaf243787f85a2ef9213ccf13ab4a243283fde (diff) | |
| download | linux-8a2e70c40ff58f82dde67770e6623ca45f0cb0c8.tar.bz2 | |
ocfs2: Count more refcount records in file system fragmentation.
The refcount record calculation in ocfs2_calc_refcount_meta_credits
is too optimistic that we can always allocate contiguous clusters
and handle an already existed refcount rec as a whole. Actually
because of file system fragmentation, we may have the chance to split
a refcount record into 3 parts during the transaction. So consider
the worst case in record calculation.
Cc: stable@kernel.org
Signed-off-by: Tao Ma <tao.ma@oracle.com>
Signed-off-by: Joel Becker <joel.becker@oracle.com>
Diffstat (limited to 'fs/ocfs2/refcounttree.c')
| -rw-r--r-- | fs/ocfs2/refcounttree.c | 20 |
1 files changed, 15 insertions, 5 deletions
diff --git a/fs/ocfs2/refcounttree.c b/fs/ocfs2/refcounttree.c index 3ac5aa733e9c..73a11ccfd4c2 100644 --- a/fs/ocfs2/refcounttree.c +++ b/fs/ocfs2/refcounttree.c @@ -2436,16 +2436,26 @@ static int ocfs2_calc_refcount_meta_credits(struct super_block *sb, len = min((u64)cpos + clusters, le64_to_cpu(rec.r_cpos) + le32_to_cpu(rec.r_clusters)) - cpos; /* - * If the refcount rec already exist, cool. We just need - * to check whether there is a split. Otherwise we just need - * to increase the refcount. - * If we will insert one, increases recs_add. - * * We record all the records which will be inserted to the * same refcount block, so that we can tell exactly whether * we need a new refcount block or not. + * + * If we will insert a new one, this is easy and only happens + * during adding refcounted flag to the extent, so we don't + * have a chance of spliting. We just need one record. + * + * If the refcount rec already exists, that would be a little + * complicated. we may have to: + * 1) split at the beginning if the start pos isn't aligned. + * we need 1 more record in this case. + * 2) split int the end if the end pos isn't aligned. + * we need 1 more record in this case. + * 3) split in the middle because of file system fragmentation. + * we need 2 more records in this case(we can't detect this + * beforehand, so always think of the worst case). */ if (rec.r_refcount) { + recs_add += 2; /* Check whether we need a split at the beginning. */ if (cpos == start_cpos && cpos != le64_to_cpu(rec.r_cpos)) |