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synced 2025-01-12 18:34:36 +08:00
Fix nasty performance problem in tsquery_rewrite().
tsquery_rewrite() tries to find matches to subsets of AND/OR conditions; for example, in the query 'a | b | c' the substitution subquery 'a | c' should match and lead to replacement of the first and third items. That's fine, but the matching algorithm apparently takes about O(2^N) for an N-clause query (I say "apparently" because the code is also both unintelligible and uncommented). We could probably do better than that even without any extra assumptions --- but actually, we know that the subclauses are sorted, indeed are depending on that elsewhere in this very same function. So we can just scan the two lists a single time to detect matches, as though we were doing a merge join. Also do a re-flattening call (QTNTernary()) in tsquery_rewrite_query, just to make sure that the tree fits the expectations of the next search cycle. I didn't try to devise a test case for this, but I'm pretty sure that the oversight could have led to failure to match in some cases where a match would be expected. Improve comments, and also stick a CHECK_FOR_INTERRUPTS into dofindsubquery, just in case it's still too slow for somebody. Per report from Andreas Seltenreich. Back-patch to all supported branches. Discussion: <8760oasf2y.fsf@credativ.de>
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@ -21,47 +21,43 @@
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#include "utils/builtins.h"
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static int
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addone(int *counters, int last, int total)
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{
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/* since this function recurses, it could be driven to stack overflow. */
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check_stack_depth();
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counters[last]++;
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if (counters[last] >= total)
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{
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if (last == 0)
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return 0;
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if (addone(counters, last - 1, total - 1) == 0)
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return 0;
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counters[last] = counters[last - 1] + 1;
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}
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return 1;
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}
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/*
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* If node is equal to ex, replace it with subs. Replacement is actually done
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* by returning either node or a copy of subs.
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* If "node" is equal to "ex", return a copy of "subs" instead.
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* If "ex" matches a subset of node's children, return a modified version
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* of "node" in which those children are replaced with a copy of "subs".
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* Otherwise return "node" unmodified.
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*
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* The QTN_NOCHANGE bit is set in successfully modified nodes, so that
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* we won't uselessly recurse into them.
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* Also, set *isfind true if we make a replacement.
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*/
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static QTNode *
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findeq(QTNode *node, QTNode *ex, QTNode *subs, bool *isfind)
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{
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/* Can't match unless signature matches and node type matches. */
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if ((node->sign & ex->sign) != ex->sign ||
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node->valnode->type != ex->valnode->type)
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return node;
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/* Ignore nodes marked NOCHANGE, too. */
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if (node->flags & QTN_NOCHANGE)
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return node;
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if (node->valnode->type == QI_OPR)
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{
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/* Must be same operator. */
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if (node->valnode->qoperator.oper != ex->valnode->qoperator.oper)
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return node;
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if (node->nchild == ex->nchild)
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{
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/*
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* Simple case: when same number of children, match if equal.
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* (This is reliable when the children were sorted earlier.)
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*/
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if (QTNEq(node, ex))
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{
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/* Match; delete node and return a copy of subs instead. */
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QTNFree(node);
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if (subs)
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{
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@ -73,79 +69,92 @@ findeq(QTNode *node, QTNode *ex, QTNode *subs, bool *isfind)
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*isfind = true;
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}
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}
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else if (node->nchild > ex->nchild)
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else if (node->nchild > ex->nchild && ex->nchild > 0)
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{
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/*
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* AND and NOT are commutative, so we check if a subset of the
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* children match. For example, if tnode is A | B | C, and ex is B
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* | C, we have a match after we convert tnode to A | (B | C).
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* AND and OR are commutative/associative, so we should check if a
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* subset of the children match. For example, if node is A|B|C,
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* and ex is B|C, we have a match after we notionally convert node
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* to A|(B|C). This does not work for NOT or PHRASE nodes, but we
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* can't get here for those node types because they have a fixed
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* number of children.
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*
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* Because we expect that the children are sorted, it suffices to
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* make one pass through the two lists to find the matches.
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*/
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int *counters = (int *) palloc(sizeof(int) * node->nchild);
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int i;
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QTNode *tnode = (QTNode *) palloc(sizeof(QTNode));
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bool *matched;
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int nmatched;
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int i,
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j;
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memset(tnode, 0, sizeof(QTNode));
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tnode->child = (QTNode **) palloc(sizeof(QTNode *) * ex->nchild);
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tnode->nchild = ex->nchild;
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tnode->valnode = (QueryItem *) palloc(sizeof(QueryItem));
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*(tnode->valnode) = *(ex->valnode);
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/* Assert that the subset rule is OK */
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Assert(node->valnode->qoperator.oper == OP_AND ||
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node->valnode->qoperator.oper == OP_OR);
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for (i = 0; i < ex->nchild; i++)
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counters[i] = i;
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do
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/* matched[] will record which children of node matched */
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matched = (bool *) palloc0(node->nchild * sizeof(bool));
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nmatched = 0;
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i = j = 0;
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while (i < node->nchild && j < ex->nchild)
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{
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tnode->sign = 0;
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for (i = 0; i < ex->nchild; i++)
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int cmp = QTNodeCompare(node->child[i], ex->child[j]);
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if (cmp == 0)
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{
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tnode->child[i] = node->child[counters[i]];
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tnode->sign |= tnode->child[i]->sign;
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/* match! */
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matched[i] = true;
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nmatched++;
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i++, j++;
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}
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if (QTNEq(tnode, ex))
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else if (cmp < 0)
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{
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int j = 0;
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pfree(tnode->valnode);
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pfree(tnode->child);
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pfree(tnode);
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if (subs)
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{
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tnode = QTNCopy(subs);
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tnode->flags = QTN_NOCHANGE | QTN_NEEDFREE;
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}
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else
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tnode = NULL;
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node->child[counters[0]] = tnode;
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for (i = 1; i < ex->nchild; i++)
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node->child[counters[i]] = NULL;
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for (i = 0; i < node->nchild; i++)
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{
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if (node->child[i])
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{
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node->child[j] = node->child[i];
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j++;
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}
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}
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node->nchild = j;
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*isfind = true;
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/* node->child[i] has no match, ignore it */
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i++;
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}
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else
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{
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/* ex->child[j] has no match; we can give up immediately */
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break;
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}
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} while (addone(counters, ex->nchild - 1, node->nchild));
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if (tnode && (tnode->flags & QTN_NOCHANGE) == 0)
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{
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pfree(tnode->valnode);
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pfree(tnode->child);
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pfree(tnode);
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}
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else
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if (nmatched == ex->nchild)
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{
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/* collapse out the matched children of node */
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j = 0;
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for (i = 0; i < node->nchild; i++)
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{
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if (matched[i])
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QTNFree(node->child[i]);
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else
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node->child[j++] = node->child[i];
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}
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/* and instead insert a copy of subs */
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if (subs)
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{
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subs = QTNCopy(subs);
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subs->flags |= QTN_NOCHANGE;
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node->child[j++] = subs;
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}
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node->nchild = j;
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/*
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* Re-sort the node to put new child in the right place. This
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* is a bit bogus, because it won't matter for findsubquery's
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* remaining processing, and it's insufficient to prepare the
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* tree for another search (we would need to re-flatten as
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* well, and we don't want to do that because we'd lose the
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* QTN_NOCHANGE marking on the new child). But it's needed to
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* keep the results the same as the regression tests expect.
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*/
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QTNSort(node);
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pfree(counters);
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*isfind = true;
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}
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pfree(matched);
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}
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}
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else
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@ -173,12 +182,20 @@ findeq(QTNode *node, QTNode *ex, QTNode *subs, bool *isfind)
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return node;
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}
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/*
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* Recursive guts of findsubquery(): attempt to replace "ex" with "subs"
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* at the root node, and if we failed to do so, recursively match against
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* child nodes.
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*/
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static QTNode *
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dofindsubquery(QTNode *root, QTNode *ex, QTNode *subs, bool *isfind)
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{
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/* since this function recurses, it could be driven to stack overflow. */
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check_stack_depth();
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/* also, since it's a bit expensive, let's check for query cancel. */
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CHECK_FOR_INTERRUPTS();
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root = findeq(root, ex, subs, isfind);
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if (root && (root->flags & QTN_NOCHANGE) == 0 && root->valnode->type == QI_OPR)
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@ -192,6 +209,10 @@ dofindsubquery(QTNode *root, QTNode *ex, QTNode *subs, bool *isfind)
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return root;
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}
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/*
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* Delete any void subtrees that may have been inserted when the replacement
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* subtree is void.
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*/
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static QTNode *
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dropvoidsubtree(QTNode *root)
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{
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@ -231,6 +252,14 @@ dropvoidsubtree(QTNode *root)
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return root;
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}
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/*
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* Substitute "subs" for "ex" throughout the QTNode tree at root.
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*
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* If isfind isn't NULL, set *isfind to show whether we made any substitution.
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*
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* Both "root" and "ex" must have been through QTNTernary and QTNSort
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* to ensure reliable matching.
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*/
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QTNode *
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findsubquery(QTNode *root, QTNode *ex, QTNode *subs, bool *isfind)
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{
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@ -344,6 +373,7 @@ tsquery_rewrite_query(PG_FUNCTION_ARGS)
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{
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/* ready the tree for another pass */
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QTNClearFlags(tree, QTN_NOCHANGE);
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QTNTernary(tree);
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QTNSort(tree);
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}
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}
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