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Retry after buffer locking failure during SPGiST index creation.
The original coding thought this case was impossible, but it can happen if the bgwriter or checkpointer processes decide to write out an index page while creation is still proceeding, leading to a bogus "unexpected spgdoinsert() failure" error. Problem reported by Jonathan S. Katz. Teodor Sigaev
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@ -1946,9 +1946,12 @@ spgdoinsert(Relation index, SpGistState *state,
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* Attempt to acquire lock on child page. We must beware of
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* deadlock against another insertion process descending from that
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* page to our parent page (see README). If we fail to get lock,
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* abandon the insertion and tell our caller to start over. XXX
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* this could be improved; perhaps it'd be worth sleeping a bit
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* before giving up?
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* abandon the insertion and tell our caller to start over.
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*
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* XXX this could be improved, because failing to get lock on a
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* buffer is not proof of a deadlock situation; the lock might be
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* held by a reader, or even just background writer/checkpointer
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* process. Perhaps it'd be worth retrying after sleeping a bit?
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*/
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if (!ConditionalLockBuffer(current.buffer))
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{
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@ -45,10 +45,17 @@ spgistBuildCallback(Relation index, HeapTuple htup, Datum *values,
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/* Work in temp context, and reset it after each tuple */
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oldCtx = MemoryContextSwitchTo(buildstate->tmpCtx);
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/* No concurrent insertions can be happening, so failure is unexpected */
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if (!spgdoinsert(index, &buildstate->spgstate, &htup->t_self,
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*values, *isnull))
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elog(ERROR, "unexpected spgdoinsert() failure");
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/*
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* Even though no concurrent insertions can be happening, we still might
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* get a buffer-locking failure due to bgwriter or checkpointer taking a
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* lock on some buffer. So we need to be willing to retry. We can flush
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* any temp data when retrying.
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*/
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while (!spgdoinsert(index, &buildstate->spgstate, &htup->t_self,
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*values, *isnull))
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{
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MemoryContextReset(buildstate->tmpCtx);
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}
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MemoryContextSwitchTo(oldCtx);
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MemoryContextReset(buildstate->tmpCtx);
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