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>
This commit is contained in:
Tao Ma 2010-07-22 13:56:45 +08:00 коммит произвёл Joel Becker
Родитель 7beaf24378
Коммит 8a2e70c40f
1 изменённых файлов: 15 добавлений и 5 удалений

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@ -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))