Btrfs: Only let very young transactions grow during commit
Commits are fairly expensive, and so btrfs has code to sit around for a while during the commit and let new writers come in. But, while we're sitting there, new delayed refs might be added, and those can be expensive to process as well. Unless the transaction is very very young, it makes sense to go ahead and let the commit finish without hanging around. The commit grow loop isn't as important as it used to be, the fsync logging code handles most performance critical syncs now. Signed-off-by: Chris Mason <chris.mason@oracle.com>
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@ -972,6 +972,8 @@ int btrfs_commit_transaction(struct btrfs_trans_handle *trans,
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struct extent_io_tree *pinned_copy;
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DEFINE_WAIT(wait);
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int ret;
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int should_grow = 0;
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unsigned long now = get_seconds();
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/* make a pass through all the delayed refs we have so far
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* any runnings procs may add more while we are here
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@ -1029,6 +1031,9 @@ int btrfs_commit_transaction(struct btrfs_trans_handle *trans,
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}
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}
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if (now < cur_trans->start_time || now - cur_trans->start_time < 1)
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should_grow = 1;
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do {
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int snap_pending = 0;
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joined = cur_trans->num_joined;
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@ -1041,7 +1046,7 @@ int btrfs_commit_transaction(struct btrfs_trans_handle *trans,
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if (cur_trans->num_writers > 1)
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timeout = MAX_SCHEDULE_TIMEOUT;
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else
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else if (should_grow)
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timeout = 1;
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mutex_unlock(&root->fs_info->trans_mutex);
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@ -1051,12 +1056,14 @@ int btrfs_commit_transaction(struct btrfs_trans_handle *trans,
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BUG_ON(ret);
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}
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schedule_timeout(timeout);
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smp_mb();
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if (cur_trans->num_writers > 1 || should_grow)
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schedule_timeout(timeout);
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mutex_lock(&root->fs_info->trans_mutex);
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finish_wait(&cur_trans->writer_wait, &wait);
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} while (cur_trans->num_writers > 1 ||
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(cur_trans->num_joined != joined));
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(should_grow && cur_trans->num_joined != joined));
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ret = create_pending_snapshots(trans, root->fs_info);
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BUG_ON(ret);
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