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authorTao Ma <tao.ma@oracle.com>2010-07-22 01:56:45 -0400
committerJoel Becker <joel.becker@oracle.com>2010-08-07 13:44:49 -0400
commit8a2e70c40ff58f82dde67770e6623ca45f0cb0c8 (patch)
treece05111dbb00fae81d0c0b7726ddac5c0117ec34
parent7beaf243787f85a2ef9213ccf13ab4a243283fde (diff)
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>
-rw-r--r--fs/ocfs2/refcounttree.c20
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,
2436 len = min((u64)cpos + clusters, le64_to_cpu(rec.r_cpos) + 2436 len = min((u64)cpos + clusters, le64_to_cpu(rec.r_cpos) +
2437 le32_to_cpu(rec.r_clusters)) - cpos; 2437 le32_to_cpu(rec.r_clusters)) - cpos;
2438 /* 2438 /*
2439 * If the refcount rec already exist, cool. We just need
2440 * to check whether there is a split. Otherwise we just need
2441 * to increase the refcount.
2442 * If we will insert one, increases recs_add.
2443 *
2444 * We record all the records which will be inserted to the 2439 * We record all the records which will be inserted to the
2445 * same refcount block, so that we can tell exactly whether 2440 * same refcount block, so that we can tell exactly whether
2446 * we need a new refcount block or not. 2441 * we need a new refcount block or not.
2442 *
2443 * If we will insert a new one, this is easy and only happens
2444 * during adding refcounted flag to the extent, so we don't
2445 * have a chance of spliting. We just need one record.
2446 *
2447 * If the refcount rec already exists, that would be a little
2448 * complicated. we may have to:
2449 * 1) split at the beginning if the start pos isn't aligned.
2450 * we need 1 more record in this case.
2451 * 2) split int the end if the end pos isn't aligned.
2452 * we need 1 more record in this case.
2453 * 3) split in the middle because of file system fragmentation.
2454 * we need 2 more records in this case(we can't detect this
2455 * beforehand, so always think of the worst case).
2447 */ 2456 */
2448 if (rec.r_refcount) { 2457 if (rec.r_refcount) {
2458 recs_add += 2;
2449 /* Check whether we need a split at the beginning. */ 2459 /* Check whether we need a split at the beginning. */
2450 if (cpos == start_cpos && 2460 if (cpos == start_cpos &&
2451 cpos != le64_to_cpu(rec.r_cpos)) 2461 cpos != le64_to_cpu(rec.r_cpos))