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xfs: fix broken bhold behavior in xrep_roll_ag_trans
In xrep_roll_ag_trans, the transaction roll will always set sc->tp to the new transaction, even if committing the old one fails. A bare transaction roll leaves the buffer(s) locked but not joined to the new transaction, so it's not necessary to release the hold if the roll fails. Remove the incorrect xfs_trans_bhold_release calls. Signed-off-by: Darrick J. Wong <darrick.wong@oracle.com> Reviewed-by: Brian Foster <bfoster@redhat.com>
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@ -136,10 +136,16 @@ xrep_roll_ag_trans(
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if (sc->sa.agfl_bp)
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xfs_trans_bhold(sc->tp, sc->sa.agfl_bp);
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/* Roll the transaction. */
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/*
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* Roll the transaction. We still own the buffer and the buffer lock
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* regardless of whether or not the roll succeeds. If the roll fails,
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* the buffers will be released during teardown on our way out of the
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* kernel. If it succeeds, we join them to the new transaction and
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* move on.
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*/
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error = xfs_trans_roll(&sc->tp);
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if (error)
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goto out_release;
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return error;
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/* Join AG headers to the new transaction. */
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if (sc->sa.agi_bp)
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@ -150,21 +156,6 @@ xrep_roll_ag_trans(
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xfs_trans_bjoin(sc->tp, sc->sa.agfl_bp);
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return 0;
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out_release:
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/*
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* Rolling failed, so release the hold on the buffers. The
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* buffers will be released during teardown on our way out
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* of the kernel.
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*/
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if (sc->sa.agi_bp)
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xfs_trans_bhold_release(sc->tp, sc->sa.agi_bp);
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if (sc->sa.agf_bp)
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xfs_trans_bhold_release(sc->tp, sc->sa.agf_bp);
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if (sc->sa.agfl_bp)
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xfs_trans_bhold_release(sc->tp, sc->sa.agfl_bp);
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return error;
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}
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/*
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