黄东旭解析 TiDB 的核心优势
470
2024-02-27
这篇文章主要介绍“***中使用动态规划算法构造连接路径的实现函数是哪个”,在日常操作中,相信很多人在***中使用动态规划算法构造连接路径的实现函数是哪个问题上存在疑惑,小编查阅了各式资料,整理出简单好用的操作方法,希望对大家解答”***中使用动态规划算法构造连接路径的实现函数是哪个”的疑惑有所帮助!接下来,请跟着小编一起来学习吧!
上节已解读了make_rel_from_joinlist->standard_join_search函数的主实现逻辑,下面重点介绍该函数中的join_search_one_level函数.
/* * join_search_one_level * Consider ways to produce join relations containing exactly level * jointree items. (This is one step of the dynamic-programming method * embodied in standard_join_search.) Join rel nodes for each feasible * combination of lower-level rels are created and returned in a list. * Implementation paths are created for each such joinrel, too. * 规划如何生成包含匹配Leve(比如2个关系的连接/3个关系的连接等)连接关系。 * (这是在standard_join_search中体现的动态规划算法的一个步骤。) * 为较低Leve的关系创建新的连接关系亦即访问路径,通过链表的方式返回(root->join_rel_level)。 * * level: level of rels we want to make this time * root->join_rel_level[j], 1 <= j < level, is a list of rels containing j items * level:关系的level,比如是2个关系还是3个关系的连接 * * The result is returned in root->join_rel_level[level]. * 结果通过root->join_rel_level[level] */void join_search_one_level(PlannerInfo *root, int level) {List**joinrels = root->join_rel_level; ListCell *r; int k; Assert(joinrels[level] == NIL);/* Set join_cur_level so that new joinrels are added to proper list */ root->join_cur_level = level;//当前的Level /* * First, consider left-sided and right-sided plans, in which rels of * exactly level-1 member relations are joined against initial relations. * We prefer to join using join clauses, but if we find a rel of level-1 * members that has no join clauses, we will generate Cartesian-product * joins against all initial rels not already contained in it. * 首先,规划left-sided和right-sided的计划,这些计划已由初始关系连接为level-1级的Relation. * PG使用连接条件进行连接,但如果发现level-1成员中没有连接条件,那么PG将会 * 为未包含此条件的初始关系生成笛卡尔积. */ foreach(r, joinrels[level - 1])//遍历上一级生成的关系{ RelOptInfo *old_rel = (RelOptInfo *) lfirst(r);//获取上一级的RelOptInfo if(old_rel->joininfo != NIL || old_rel->has_eclass_joins || has_join_restriction(root, old_rel))//存在连接条件 { /* * There are join clauses or join order restrictions relevant to * this rel, so consider joins between this rel and (only) those * initial rels it is linked to by a clause or restriction. * 存在与此rel相关的连接条件或连接顺序限制, * 因此仅规划此rel与通过条件子句或约束条件链接在一起的初始rels. * * At level 2 this condition is symmetric, so there is no need to * look at initial rels before this one in the list; we already * considered such joins when we were at the earlier rel. (The * mirror-image joins are handled automatically by make_join_rel.) * In later passes (level > 2), we join rels of the previous level * to each initial rel they dont already include but have a join * clause or restriction with. * leve=2时,这个条件是对称的,所以不需要在关注链表中此rel前的rels; * 在处理在此rel前的rels时,已处理这样的连接.(make_join_rel函数自动处理镜像连接)。 * level>2时,PG将上一级别生成的rels逐一与尚未处理的初始rel(存在连接条件或约束条件)进行连接. * */ListCell *other_rels;if (level == 2) /* consider remaining initial rels */ other_rels = lnext(r);//level = 2,只需关注此rel之后的rel else /* consider all initial rels */ other_rels = list_head(joinrels[1]);//level > 2,从第1级开始尝试make_rels_by_clause_joins(root, old_rel, other_rels);//创建连接 } else//不存在连接条件 { /* * Oops, we have a relation that is not joined to any other * relation, either directly or by join-order restrictions. * Cartesian product time. * 有一个relation与其他relation没有连接条件(直接或通过join-order约束) * 笛卡尔时间到了! * * We consider a cartesian product with each not-already-included * initial rel, whether it has other join clauses or not. At * level 2, if there are two or more clauseless initial rels, we * will redundantly consider joining them in both directions; but * such cases arent common enough to justify adding complexity to * avoid the duplicated effort. * 考察每一个尚未处理的初始rel(无论其是否有约束条件). * 在level 2,如存在2个或以上的无条件初始rels,PG可能会出现重复处理的情况. */make_rels_by_clauseless_joins(root, old_rel, list_head(joinrels[1]));//创建无条件连接 } } /* * Now, consider "bushy plans" in which relations of k initial rels are * joined to relations of level-k initial rels, for 2 <= k <= level-2. * 现在考察"稠密计划",其中k level的rels与level - k的rel想连接.其中:2 <= k <= level-2 * * We only consider bushy-plan joins for pairs of rels where there is a * suitable join clause (or join order restriction), in order to avoid * unreasonable growth of planning time. * 这里只考虑存在连接条件(或者join-order限制)的关系对,以避免计划时间的大幅增加 */ for (k = 2;; k++) { int other_level = level - k;/* * Since make_join_rel(x, y) handles both x,y and y,x cases, we only * need to go as far as the halfway point. */ if (k > other_level) break; foreach(r, joinrels[k]) { RelOptInfo *old_rel = (RelOptInfo *) lfirst(r); ListCell *other_rels; ListCell *r2;/* * We can ignore relations without join clauses here, unless they * participate in join-order restrictions --- then we might have * to force a bushy join plan. */ if(old_rel->joininfo == NIL && !old_rel->has_eclass_joins && !has_join_restriction(root, old_rel))continue; if(k == other_level) other_rels = lnext(r);/*同一层次,只考虑余下的rel,only consider remaining rels */ elseother_rels = list_head(joinrels[other_level]);//不同层次,尝试所有的for_each_cell(r2, other_rels) { RelOptInfo *new_rel = (RelOptInfo *) lfirst(r2);if(!bms_overlap(old_rel->relids, new_rel->relids))//relids不存在包含关系 { /* * OK, we can build a rel of the right level from this * pair of rels. Do so if there is at least one relevant * join clause or join order restriction. */ if(have_relevant_joinclause(root, old_rel, new_rel) || have_join_order_restriction(root, old_rel, new_rel))//存在连接条件或者join-order约束{ (void) make_join_rel(root, old_rel, new_rel);//创建连接} } } } }/*---------- * Last-ditch effort: if we failed to find any usable joins so far, force * a set of cartesian-product joins to be generated. This handles the * special case where all the available rels have join clauses but we * cannot use any of those clauses yet. This can only happen when we are * considering a join sub-problem (a sub-joinlist) and all the rels in the * sub-problem have only join clauses with rels outside the sub-problem. * An example is * * SELECT ... FROM a INNER JOIN b ON TRUE, c, d, ... * WHERE a.w = c.x and b.y = d.z; * * If the "a INNER JOIN b" sub-problem does not get flattened into the * upper level, we must be willing to make a cartesian join of a and b; * but the code above will not have done so, because it thought that both * a and b have joinclauses. We consider only left-sided and right-sided * cartesian joins in this case (no bushy). *---------- */ if(joinrels[level] == NIL) {/* * This loop is just like the first one, except we always call * make_rels_by_clauseless_joins(). */ foreach(r, joinrels[level - 1]) { RelOptInfo *old_rel = (RelOptInfo *) lfirst(r); make_rels_by_clauseless_joins(root, old_rel, list_head(joinrels[1])); } /*---------- * When special joins are involved, there may be no legal way * to make an N-way join for some values of N. For example consider * * SELECT ... FROM t1 WHERE * x IN (SELECT ... FROM t2,t3 WHERE ...) AND * y IN (SELECT ... FROM t4,t5 WHERE ...) * * We will flatten this query to a 5-way join problem, but there are * no 4-way joins that join_is_legal() will consider legal. We have * to accept failure at level 4 and go on to discover a workable * bushy plan at level 5. * * However, if there are no special joins and no lateral references * then join_is_legal() should never fail, and so the following sanity * check is useful. *---------- */ if(joinrels[level] == NIL && root->join_info_list == NIL && !root->hasLateralRTEs) elog(ERROR,"failed to build any %d-way joins", level); } }//------------------------------------------------------------------- has_join_restriction /* * has_join_restriction * Detect whether the specified relation has join-order restrictions, * due to being inside an outer join or an IN (sub-SELECT), * or participating in any LATERAL references or multi-rel PHVs. * 判断传入的relation是否含有join-order限制条件.存在于外连接/IN(sub-SELECT)子查询/LATERAL依赖/多关系PHVs * * Essentially, this tests whether have_join_order_restriction() could * succeed with this rel and some other one. Its OK if we sometimes * say "true" incorrectly. (Therefore, we dont bother with the relatively * expensive has_legal_joinclause test.) */ staticbool has_join_restriction(PlannerInfo *root, RelOptInfo *rel) { ListCell *l;if (rel->lateral_relids != NULL || rel->lateral_referencers != NULL) return true;//存在lateral foreach(l, root->placeholder_list) { PlaceHolderInfo *phinfo = (PlaceHolderInfo *) lfirst(l);if(bms_is_subset(rel->relids, phinfo->ph_eval_at) && !bms_equal(rel->relids, phinfo->ph_eval_at))return true;//PHVs } foreach(l, root->join_info_list) { SpecialJoinInfo *sjinfo = (SpecialJoinInfo *) lfirst(l);/* ignore full joins --- other mechanisms preserve their ordering */ if(sjinfo->jointype == JOIN_FULL)continue;//不考虑全外连接 /* ignore if SJ is already contained in rel */ if(bms_is_subset(sjinfo->min_lefthand, rel->relids) && bms_is_subset(sjinfo->min_righthand, rel->relids))continue;//SJ在rel中,不考虑 /* restricted if it overlaps LHS or RHS, but doesnt contain SJ */ if(bms_overlap(sjinfo->min_lefthand, rel->relids) || bms_overlap(sjinfo->min_righthand, rel->relids))return true; } return false; } //------------------------------------------------------------------- make_rels_by_clause_joins /* * make_rels_by_clause_joins * Build joins between the given relation old_rel and other relations * that participate in join clauses that old_rel also participates in * (or participate in join-order restrictions with it). * The join rels are returned in root->join_rel_level[join_cur_level]. * 创建old_rel和其他rel的连接(两者存在连接条件) * *Note:at levels above 2 we will generate the same joined relation in * multiple ways --- for example (a join b) join c is the same RelOptInfo as * (b join c) join a, though the second case will add a different set of Paths * to it. This is the reason for using the join_rel_level mechanism, which * automatically ensures that each new joinrel is only added to the list once. * 注意:在level > 2时,PG会通过多种方式生成同样的连接rel(joined relation). * 比如:(a join b) join c与(b join c) join a最终结果是一样的RelOptInfo,虽然第 * 2种方法会添加一些不同的访问路径集合在其中. * 这其实是使用join_rel_level的原因,确保每个新joinrel只加入到合适的链表中 * * old_rel is the relation entry for the relation to be joined * other_rels: the first cell in a linked list containing the other * rels to be considered for joining * old-rel:需要连接的rel * other-rel:候选关系链表中的的第一个cell * * Currently, this is only used with initial rels in other_rels, but it * will work for joining to joinrels too. * 看起来似乎只对other_rels中的初始rels有用,但其实对于连接生成的joinrels同样会生效. */ staticvoid make_rels_by_clause_joins(PlannerInfo *root, RelOptInfo *old_rel, ListCell *other_rels) { ListCell *l; for_each_cell(l, other_rels)//遍历链表{ RelOptInfo *other_rel = (RelOptInfo *) lfirst(l);//获取其中的RelOptInfo if(!bms_overlap(old_rel->relids, other_rel->relids) && (have_relevant_joinclause(root, old_rel, other_rel) || have_join_order_restriction(root, old_rel, other_rel)))//reldis不同而且存在连接关系&连接顺序约束 { (void) make_join_rel(root, old_rel, other_rel);//创建连接} } }//---------------------------------------------------- have_relevant_joinclause /* * have_relevant_joinclause * Detect whether there is a joinclause that involves * the two given relations. * 给定两个relations,检查两者是否存在连接条件 * *Note:the joinclause does not have to be evaluable with only these two * relations. This is intentional. For example consider * SELECT * FROM a, b, c WHERE a.x = (b.y + c.z) * If a is much larger than the other tables, it may be worthwhile to * cross-join b and c and then use an inner indexscan on a.x. Therefore * we should consider this joinclause as reason to join b to c, even though * it cant be applied at that join step. * 注意:连接条件不一定是等值连接, * 比如:SELECT * FROM a, b, c WHERE a.x = (b.y + c.z),只要a.x大于b.y + c.z即可 */bool have_relevant_joinclause(PlannerInfo *root, RelOptInfo *rel1, RelOptInfo *rel2) { bool result =false; List*joininfo; Relids other_relids; ListCell *l;/* * We could scan either relations joininfo list; may as well use the * shorter one. * 获取relation中joininfo链表较少的那个 */ if(list_length(rel1->joininfo) <= list_length(rel2->joininfo)) { joininfo = rel1->joininfo; other_relids = rel2->relids; }else{ joininfo = rel2->joininfo; other_relids = rel1->relids; }foreach(l, joininfo)//遍历{ RestrictInfo *rinfo = (RestrictInfo *) lfirst(l);if (bms_overlap(other_relids, rinfo->required_relids))//存在交集{ result =true;//存在连接条件 break; } } /* * We also need to check the EquivalenceClass data structure, which might * contain relationships not emitted into the joininfo lists. * 检查等价类 */ if(!result && rel1->has_eclass_joins && rel2->has_eclass_joins) result = have_relevant_eclass_joinclause(root, rel1, rel2);//存在等价类连接条件 return result; } //---------------------------------------------------- have_join_order_restriction /* * have_join_order_restriction * Detect whether the two relations should be joined to satisfy * a join-order restriction arising from special or lateral joins. * 检查两个relations是否需要连接以满足join-order限制(由于special/lateral连接引起) * * In practice this is always used with have_relevant_joinclause(), and so * could be merged with that function, but it seems clearer to separate the * two concerns. We need this test because there are degenerate cases where * a clauseless join must be performed to satisfy join-order restrictions. * Also, if one rel has a lateral reference to the other, or both are needed * to compute some PHV, we should consider joining them even if the join would * be clauseless. * 在实践中,这通常与have_relevance _join子()一起使用,因此可以与该函数合并, * 但分离这两个关注点似乎更为清晰。在一些退化的情况下需要这个测试, * 必须执行无语法连接以满足连接顺序限制。 * 另外,如果一个rel与另一个rel有一个lateral引用, * 或者两者都需要计算一些PHV,那么我们应该考虑加入它们,即使连接是无连接条件的。 * *Note:this is only a problem if one side of a degenerate outer join * contains multiple rels, or a clauseless join is required within an * IN/EXISTS RHS; else we will find a join path via the "last ditch" case in * join_search_one_level(). We could dispense with this test if we were * willing to try bushy plans in the "last ditch" case, but that seems much * less efficient. * 注意:只有当简并外部连接的一侧包含多个rels时, * 或者在IN/EXISTS RHS中需要一个无修饰的连接时,才会出现这个问题; * 否则,将通过join_search_one_level()中的“last ditch” * 找到连接路径。如果愿意在“稠密计划”的情况下进行大量的尝试, * 那么可以省去这个测试,但这似乎效率要低得多。 */bool have_join_order_restriction(PlannerInfo *root, RelOptInfo *rel1, RelOptInfo *rel2) { bool result =false; ListCell *l;/* * If either side has a direct lateral reference to the other, attempt the * join regardless of outer-join considerations. */ if(bms_overlap(rel1->relids, rel2->direct_lateral_relids) || bms_overlap(rel2->relids, rel1->direct_lateral_relids))return true;//relids与lateral relids存在交集,返回T /* * Likewise, if both rels are needed to compute some PlaceHolderVar, * attempt the join regardless of outer-join considerations. (This is not * very desirable, because a PHV with a large eval_at set will cause a lot * of probably-useless joins to be considered, but failing to do this can * cause us to fail to construct a plan at all.) */ foreach(l, root->placeholder_list)//遍历PHV{ PlaceHolderInfo *phinfo = (PlaceHolderInfo *) lfirst(l);if(bms_is_subset(rel1->relids, phinfo->ph_eval_at) && bms_is_subset(rel2->relids, phinfo->ph_eval_at))return true; } /* * Its possible that the rels correspond to the left and right sides of a * degenerate outer join, that is, one with no joinclause mentioning the * non-nullable side; in which case we should force the join to occur. * * Also, the two rels could represent a clauseless join that has to be * completed to build up the LHS or RHS of an outer join. */ foreach(l, root->join_info_list)//遍历连接链表 { SpecialJoinInfo *sjinfo = (SpecialJoinInfo *) lfirst(l); /* ignore full joins --- other mechanisms handle them */ if (sjinfo->jointype == JOIN_FULL) continue; /* Can we perform the SJ with these rels? */ if(bms_is_subset(sjinfo->min_lefthand, rel1->relids) && bms_is_subset(sjinfo->min_righthand, rel2->relids)) { result =true; break; } if(bms_is_subset(sjinfo->min_lefthand, rel2->relids) && bms_is_subset(sjinfo->min_righthand, rel1->relids)) { result =true; break; } /* * Might we need to join these rels to complete the RHS? We have to * use "overlap" tests since either rel might include a lower SJ that * has been proven to commute with this one. */ if(bms_overlap(sjinfo->min_righthand, rel1->relids) && bms_overlap(sjinfo->min_righthand, rel2->relids)) { result =true; break; } /* Likewise for the LHS. */ if(bms_overlap(sjinfo->min_lefthand, rel1->relids) && bms_overlap(sjinfo->min_lefthand, rel2->relids)) { result =true; break; } } /* * We do not force the join to occur if either input rel can legally be * joined to anything else using joinclauses. This essentially means that * clauseless bushy joins are put off as long as possible. The reason is * that when there is a join order restriction high up in the join tree * (that is, with many rels inside the LHS or RHS), we would otherwise * expend lots of effort considering very stupid join combinations within * its LHS or RHS. */ if (result) { if(has_legal_joinclause(root, rel1) || has_legal_joinclause(root, rel2)) result =false; } return result; }二、跟踪分析创建测试数据表并生成测试数据:
drop table if exists a; drop table if exists b; drop table if exists c; drop table if exists d; drop table if exists e; drop table if exists f; create table a(c1 int,c2 varchar(20)); create table b(c1 int,c2 varchar(20)); create table c(c1 int,c2 varchar(20)); create table d(c1 int,c2 varchar(20)); create table e(c1 int,c2 varchar(20)); create table f(c1 int,c2 varchar(20)); insert into a select generate_series(1,100),TEST||generate_series(1,100); insert into b select generate_series(1,1000),TEST||generate_series(1,1000); insert into c select generate_series(1,10000),TEST||generate_series(1,10000); insert into d select generate_series(1,200),TEST||generate_series(1,200); insert into e selectgenerate_series(1,4000),TEST||generate_series(1,4000); insert into f select generate_series(1,100000),TEST||generate_series(1,100000);测试脚本:
testdb=# explain verbose select a.*,b.c1,c.c2,d.c2,e.c1,f.c2 froma inner join bon a.c1=b.c1,c,d,e inner join f on e.c1 = f.c1 and e.c1 < 100 where a.c1=f.c1 and b.c1=c.c1 andc.c1 = d.c1andd.c1 = e.c1; QUERY PLAN ---------------------------------------------------------------------------------------------------------- Nested Loop (cost=101.17..2218.24 rows=2width=42) Output: a.c1, a.c2, b.c1, c.c2, d.c2, e.c1, f.c2 Join Filter: (a.c1 = b.c1) -> Hash Join (cost=3.25..196.75 rows=100 width=22) Output: a.c1, a.c2, c.c2, c.c1 Hash Cond:(c.c1 = a.c1) -> Seq Scan on public.c (cost=0.00..155.00 rows=10000 width=12) Output: c.c1, c.c2 -> Hash (cost=2.00..2.00 rows=100width=10) Output: a.c1, a.c2 -> Seq Scan onpublic.a (cost=0.00..2.00 rows=100 width=10) Output: a.c1, a.c2 -> Materialize (cost=97.92..2014.00 rows=5 width=32) Output: b.c1, d.c2, d.c1, e.c1, f.c2, f.c1 -> Hash Join (cost=97.92..2013.97 rows=5 width=32) Output: b.c1, d.c2, d.c1, e.c1, f.c2, f.c1 Hash Cond:(f.c1 = b.c1) -> Seq Scan onpublic.f (cost=0.00..1541.00 rows=100000 width=13) Output: f.c1, f.c2 -> Hash (cost=97.86..97.86 rows=5 width=19) Output: b.c1, d.c2, d.c1, e.c1 -> Hash Join (cost=78.10..97.86 rows=5 width=19) Output: b.c1, d.c2, d.c1, e.c1 Hash Cond:(b.c1 = e.c1) -> Seq Scan on public.b (cost=0.00..16.00 rows=1000 width=4) Output: b.c1, b.c2 -> Hash (cost=78.04..78.04rows=5 width=15) Output: d.c2, d.c1, e.c1 -> Hash Join (cost=73.24..78.04 rows=5 width=15) Output: d.c2, d.c1, e.c1 Hash Cond:(d.c1 = e.c1) -> Seq Scan on public.d (cost=0.00..4.00 rows=200 width=11) Output: d.c1, d.c2 -> Hash (cost=72.00..72.00 rows=99 width=4) Output: e.c1 -> Seq Scanon public.e (cost=0.00..72.00 rows=99width=4) Output: e.c1 Filter: (e.c1 <100) (38 rows)测试SQL语句的连接关系:a-b,a-f,b-c,c-d,d-e,e-f
注:根据先前章节的知识,该SQL语句存在等价类{a.c1 b.c1 c.c1 d.c1 e.c1 f.c1}启动gdb跟踪
(gdb) b join_search_one_level Breakpoint 1 at 0x755667: file joinrels.c, line 67. (gdb) c Continuing. Breakpoint 1, join_search_one_level (root=0x3006e28, level=2) at joinrels.c:67 67 List **joinrels = root->join_rel_level;查看优化器信息(root)
(gdb) p *root $13 = {type = T_PlannerInfo, parse = 0x2fa3410, glob = 0x3008578, query_level = 1, parent_root = 0x0, plan_params = 0x0, outer_params = 0x0, simple_rel_array = 0x2f510e8, simple_rel_array_size = 9, simple_rte_array = 0x2f51178, all_baserels = 0x2f53dd8, nullable_baserels = 0x0, join_rel_list = 0x2fcb5c8, join_rel_hash = 0x0, join_rel_level = 0x2fcafe8, join_cur_level = 2, init_plans = 0x0, cte_plan_ids = 0x0, multiexpr_params = 0x0, eq_classes = 0x2f52cb8, canon_pathkeys = 0x2fcb718, left_join_clauses = 0x0, right_join_clauses = 0x0, full_join_clauses = 0x0, join_info_list = 0x0, append_rel_list = 0x0, rowMarks = 0x0, placeholder_list = 0x0, fkey_list = 0x0, query_pathkeys = 0x0, group_pathkeys = 0x0, window_pathkeys = 0x0, distinct_pathkeys = 0x0, sort_pathkeys = 0x0, part_schemes = 0x0, initial_rels = 0x2fcaf18, upper_rels = {0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0}, upper_targets = {0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0}, processed_tlist = 0x2f4f718, grouping_map = 0x0, minmax_aggs = 0x0, planner_cxt = 0x2e87040, total_table_pages = 627, tuple_fraction = 0, limit_tuples = -1, qual_security_level = 0, inhTargetKind = INHKIND_NONE, hasJoinRTEs = true, hasLateralRTEs = false, hasDeletedRTEs = false, hasHavingQual = false, hasPseudoConstantQuals = false, hasRecursion = false, wt_param_id = -1, non_recursive_path = 0x0, curOuterRels = 0x0, curOuterParams = 0x0, join_search_private = 0x0, partColsUpdated = false}root->simple_rel_array_size=9,数组中有9个元素,从1-8(下标为0的元素无用)分别是1->RTE_RELATION/16775,2->RTE_RELATION/16778,3->RTE_JOIN,4->RTE_RELATION/16781,5->RTE_RELATION/16784,6->RTE_RELATION/16787,7->RTE_RELATION/16790,8->RTE_JOIN
oid | relname -------+--------- 16775 | a -->1 16778 | b -->2 16781| c-->4 16784 | d -->5 16787 | e -->6 16790 | f -->7 (6 rows)进入join_search_one_level函数,level=2,开始循环遍历joinrels
(gdb) n 74 root->join_cur_level = level; (gdb) 83 foreach(r, joinrels[level - 1]) (gdb) n 85 RelOptInfo *old_rel = (RelOptInfo *) lfirst(r); (gdb) 87 if (old_rel->joininfo != NIL || old_rel->has_eclass_joins || (gdb) 105 if (level == 2) /* consider remaining initial rels */ (gdb) 106 other_rels = lnext(r); (gdb) 110 make_rels_by_clause_joins(root,[level=2]进入make_rels_by_clause_joins函数
(gdb) step make_rels_by_clause_joins (root=0x3006e28, old_rel=0x3008258, other_rels=0x2fcaf48) at joinrels.c:280 280 for_each_cell(l, other_rels)[level=2]由于存在等价类{a.c1 b.c1 c.c1 d.c1 e.c1 f.c1},因此这一步骤会两两连接构造新的关系,ab,ac,ad,ae,af,bc,bd,...
(gdb) n 282 RelOptInfo *other_rel = (RelOptInfo *) lfirst(l); (gdb) 284 if (!bms_overlap(old_rel->relids, other_rel->relids) && (gdb) 285 (have_relevant_joinclause(root, old_rel, other_rel) || (gdb) 284 if (!bms_overlap(old_rel->relids, other_rel->relids) && (gdb) 288 (void) make_join_rel(root, old_rel, other_rel); (gdb) n 280 for_each_cell(l, other_rels)[level=2]调用make_join_rel函数后,查看root->join_rel_level[2],relids=6=2+4,这是1号(关系a)和2号(关系b)RTE的连接.
(gdb) p *root->join_rel_level[2] $6 = {type = T_List, length = 1, head = 0x2fcb5f8, tail = 0x2fcb5f8} (gdb) p *(Node *)root->join_rel_level[2]->head->data.ptr_value $7 = {type = T_RelOptInfo} (gdb) p *(RelOptInfo *)root->join_rel_level[2]->head->data.ptr_value $8 = {type = T_RelOptInfo, reloptkind = RELOPT_JOINREL, relids = 0x2fcb050, rows = 100, consider_startup = false, consider_param_startup = false, consider_parallel = true, reltarget = 0x2fcb068, pathlist = 0x2fcba08, ppilist = 0x0, partial_pathlist = 0x0, cheapest_startup_path = 0x0, cheapest_total_path = 0x0, cheapest_unique_path = 0x0, cheapest_parameterized_paths = 0x0, direct_lateral_relids = 0x0, lateral_relids = 0x0, relid = 0, reltablespace = 0, rtekind = RTE_JOIN, min_attr = 0, max_attr = 0, attr_needed = 0x0, attr_widths = 0x0, lateral_vars = 0x0, lateral_referencers = 0x0, indexlist = 0x0, statlist = 0x0, pages = 0, tuples = 0, allvisfrac = 0, subroot = 0x0, subplan_params = 0x0, rel_parallel_workers = -1, serverid = 0, userid = 0, useridiscurrent = false, fdwroutine = 0x0, fdw_private = 0x0, unique_for_rels = 0x0, non_unique_for_rels = 0x0, baserestrictinfo = 0x0, baserestrictcost = { startup = 0, per_tuple = 0}, baserestrict_min_security = 4294967295, joininfo = 0x0, has_eclass_joins = true, top_parent_relids = 0x0, part_scheme = 0x0, nparts = 0, boundinfo = 0x0, partition_qual = 0x0, part_rels = 0x0, partexprs = 0x0, nullable_partexprs = 0x0, partitioned_child_rels = 0x0} (gdb) set $tmp=(RelOptInfo *)root->join_rel_level[2]->head->data.ptr_value (gdb) p *$tmp->relids->words $10 = 6[level=2]继续循环,下几组分别是ac,ad,ae,af
(gdb) p *$tmp->relids->words $12 = 18/34/66/130[level=2]完成对关系a的两两连接
(gdb) n 291 } (gdb) join_search_one_level (root=0x3006e28, level=2) at joinrels.c:89 89 { (gdb) n 83 foreach(r, joinrels[level - 1])[level=2]类似的,处理b/c/d/e/f,两两形成连接,一共有15种组合(6!/(2!*(6-2)!))
(gdb) 83 foreach(r, joinrels[level - 1]) (gdb) 142 for (k = 2;; k++) (gdb) p *root->join_rel_level[2] $44 = {type = T_List, length = 15, head = 0x2fcb5f8, tail = 0x2fd7f78}[level=2]完成level=2的调用,level2的relids组合有1&2,1&4,1&5,1&6,1&7,2&4,2&5,2&6,2&7,4&5,4&6,4&7,5&6,5&7,6&7
(gdb) standard_join_search (root=0x3006e28, levels_needed=6, initial_rels=0x2fcaf18) at allpaths.c:2757 2757 foreach(lc, root->join_rel_level[lev])开始level=3的调用
(gdb) c Continuing. Breakpoint 1, join_search_one_level (root=0x3006e28, level=3) at joinrels.c:67 67 List **joinrels = root->join_rel_level;[level=3]遍历level=2的RelOptInfo(两两连接形成的新关系)
(gdb) 83 foreach(r, joinrels[level - 1])[level=3]与level=2不同,选择初始的RelOptInfo进行连接,而不是同级的rels
... (gdb) 108 other_rels = list_head(joinrels[1]);[level=3]完成第一轮的循环,root->join_rel_level[3]链表中有4个Node(RelOptInfo),其relids分别是22/38/70/134,即1&2&4,1&2&5,1&2&6,1&2&7
(gdb) p *((RelOptInfo *)root->join_rel_level[3]->head->data.ptr_value)->relids->words $55 = 22 (gdb) p *((RelOptInfo *)root->join_rel_level[3]->head->next->data.ptr_value)->relids->words $56 = 38 (gdb) p *((RelOptInfo *)root->join_rel_level[3]->head->next->next->data.ptr_value)->relids->words $57 = 70(gdb) p *((RelOptInfo *)root->join_rel_level[3]->head->next->next->next->data.ptr_value)->relids->words $58 = 134[level=3]完成所有循环后的root->join_rel_level[3],构成连接的relids组合,一共20个(请参照数学组合的计算),包括1&2&4,1&2&5,1&2&6,1&2&7,1&4&5,1&4&6,1&4&7,...
... (gdb) p *root->join_rel_level[3] $68 = {type = T_List, length = 20, head = 0x2fd90d8, tail = 0x2f7f248}[level=3]尝试bushy plans,达不到要求,退出循环
142 for (k = 2;; k++) (gdb) 144 int other_level = level - k; (gdb) 150 if (k > other_level) 150 if (k > other_level) (gdb) n 151 break;[level=3]完成level=3的调用,开始level 4调用
(gdb) standard_join_search (root=0x3006e28, levels_needed=6, initial_rels=0x2fcaf18) at allpaths.c:2757 2757 foreach(lc, root->join_rel_level[lev]) (gdb) c Continuing. Breakpoint1, join_search_one_level (root=0x3006e28, level=4) at joinrels.c:67 67 List **joinrels = root->join_rel_level;[level=4]完成第一轮循环调用,查看root->join_rel_level[4],relids分别是54/86/150,即1&2&4&5,1&2&4&6,1&2&4&7
... 89 { (gdb) 83 foreach(r, joinrels[level - 1]) (gdb) p *root->join_rel_level[4] $69= {type = T_List, length =3, head = 0x2f838e0, tail = 0x30654d8} (gdb) p *((RelOptInfo *)root->join_rel_level[4]->head->data.ptr_value)->relids->words $70 = 54 (gdb) p *((RelOptInfo *)root->join_rel_level[4]->head->next->data.ptr_value)->relids->words $71 = 86(gdb) p *((RelOptInfo *)root->join_rel_level[4]->head->next->next->data.ptr_value)->relids->words $72 = 150[level=4]所有循环后的root->join_rel_level[4],构成连接的relids组合,一共15个
(gdb) b joinrels.c:142 Breakpoint 2 at 0x75576a: file joinrels.c, line 142. (gdb) c Continuing. Breakpoint 2, join_search_one_level (root=0x3006e28, level=4) at joinrels.c:142 142 for (k = 2;; k++) (gdb) p *root->join_rel_level[4] $73 = {type = T_List, length = 15, head = 0x2f838e0, tail = 0x307bd78}[level=4]尝试bushy plans
... (gdb) p k $74 = 2 (gdb) p other_level $75 = 2[level=4]遍历k级关系,k=other_level,同一层次的rel,两两组合,即1&2,3&4等尝试两两配对连接
(gdb) n 153 foreach(r, joinrels[k]) ... (gdb) 168 if (k == other_level)[level=4]如relids=6和relids=48的两个关系
177 if (!bms_overlap(old_rel->relids, new_rel->relids)) (gdb) 184 if(have_relevant_joinclause(root, old_rel, new_rel) || (gdb) p *old_rel->relids->words $78 = 6 (gdb) p *new_rel->relids->words $79 = 48[level=4]构造新的关系,但该关系无法通过合法连接形成或者已存在,因此没有对root->join_rel_level[4]有所影响(调用前后均为15个Node)
(gdb) n 187 (void) make_join_rel(root, old_rel, new_rel); (gdb) 173 for_each_cell(r2, other_rels) (gdb) p *root->join_rel_level[4] $80 = {type = T_List, length = 15, head = 0x2f838e0, tail = 0x307bd78}[level=4]完成bushy plans,root->join_rel_level[4]元素个数没有变化
(gdb) c Continuing. Breakpoint 3, join_search_one_level (root=0x3006e28, level=4) at joinrels.c:213 213 if (joinrels[level] == NIL) (gdb) p *root->join_rel_level[4] $82 = {type = T_List, length = 15, head = 0x2f838e0, tail = 0x307bd78}[level=5]进入level=5调用
(gdb) c Continuing. Breakpoint 1, join_search_one_level (root=0x3006e28, level=5) at joinrels.c:67 67 List**joinrels = root->join_rel_level;[level=5]完成第一轮循环调用,查看root->join_rel_level[5],relids分别是118/182,即1&2&4&5&6,1&2&4&6&7
(gdb) p *root->join_rel_level[5] $83 = {type = T_List, length = 2, head = 0x30931d0, tail =0x3093dc8} (gdb) p *((RelOptInfo *)root->join_rel_level[5]->head->data.ptr_value)->relids->words $85 = 118 (gdb) p *((RelOptInfo *)root->join_rel_level[5]->head->next->data.ptr_value)->relids->words $86 = 182[level=5]所有循环后的root->join_rel_level[5],构成连接的relids组合,一共6个
(gdb) p *root->join_rel_level[5] $87 = {type = T_List, length = 6, head = 0x30931d0, tail = 0x309d188}[level=5]尝试bushy plans,即2个rels连接生成的关系 join 3个rels连接生成的关系
完成调用(gdb) c Continuing. Breakpoint 3, join_search_one_level (root=0x3006e28, level=5) at joinrels.c:213 213 if (joinrels[level] == NIL) (gdb) p *root->join_rel_level[5] $91 = {type = T_List, length = 6, head = 0x30931d0, tail = 0x309d188}[level=6]进入level=6调用
(gdb) c Continuing. Breakpoint 1, join_search_one_level (root=0x3006e28, level=6) at joinrels.c:67 67 List **joinrels = root->join_rel_level;[level=6]与level=1的rels连接后,形成1个新的关系
(gdb) c Continuing. Breakpoint 2, join_search_one_level (root=0x3006e28, level=6) at joinrels.c:142 142 for (k = 2;; k++) (gdb) p *root->join_rel_level[6] $92 = {type = T_List, length = 1, head = 0x3104cf8, tail = 0x3104cf8}[level=6]尝试bushy plans,即2个rels连接生成的关系 join 4个rels连接生成的关系 & 3 join 3
完成调用,生成level=6的结果链表(gdb) c Continuing. Breakpoint 3, join_search_one_level (root=0x3006e28, level=6) at joinrels.c:213 213 if (joinrels[level] == NIL) (gdb) p *root->join_rel_level[6] $93 = {type = T_List, length = 1, head = 0x3104cf8, tail = 0x3104cf8} (gdb) p *(RelOptInfo *)root->join_rel_level[6]->head->data.ptr_value $94 = {type = T_RelOptInfo, reloptkind = RELOPT_JOINREL, relids = 0x3099a80, rows = 2, consider_startup = false, consider_param_startup = false, consider_parallel = true, reltarget = 0x3104a08, pathlist = 0x3104ec0, ppilist = 0x0, partial_pathlist = 0x0, cheapest_startup_path = 0x0, cheapest_total_path = 0x0, cheapest_unique_path = 0x0, cheapest_parameterized_paths = 0x0, direct_lateral_relids = 0x0, lateral_relids = 0x0, relid = 0, reltablespace = 0, rtekind = RTE_JOIN, min_attr = 0, max_attr = 0, attr_needed = 0x0, attr_widths = 0x0, lateral_vars = 0x0, lateral_referencers = 0x0, indexlist = 0x0, statlist = 0x0, pages = 0, tuples = 0, allvisfrac = 0, subroot = 0x0, subplan_params = 0x0, rel_parallel_workers = -1, serverid = 0, userid = 0, useridiscurrent = false, fdwroutine = 0x0, fdw_private = 0x0, unique_for_rels = 0x0, non_unique_for_rels = 0x0, baserestrictinfo = 0x0, baserestrictcost = { startup = 0, per_tuple = 0}, baserestrict_min_security = 4294967295, joininfo = 0x0, has_eclass_joins = false, top_parent_relids = 0x0, part_scheme = 0x0, nparts = 0, boundinfo = 0x0, partit[level=6]查看访问路径
(gdb) set $roi=(RelOptInfo *)root->join_rel_level[6]->head->data.ptr_value (gdb) p *$roi->pathlist $97 = {type = T_List, length = 1, head = 0x3104ea0, tail = 0x3104ea0} (gdb) p *(Node *)$roi->pathlist->head->data.ptr_value $98 = {type = T_NestPath} (gdb) p *(NestPath *)$roi->pathlist->head->data.ptr_value $99 = {path = {type = T_NestPath, pathtype = T_NestLoop, parent = 0x31047f8, pathtarget = 0x3104a08, param_info = 0x0, parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 2, startup_cost = 101.1725, total_cost = 2218.2350000000001, pathkeys = 0x0}, jointype = JOIN_INNER, inner_unique = false, outerjoinpath = 0x2fccd80, innerjoinpath = 0x3107820, joinrestrictinfo = 0x3107ae0}该path的innerjoinpath(构造该连接inner关系的path)和outerjoinpath(构造该连接outer关系的path)
(gdb) p *$np->innerjoinpath $109 = {type = T_MaterialPath, pathtype = T_Material, parent = 0x3077c70, pathtarget = 0x3077e80, param_info = 0x0, parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 5, startup_cost = 97.922499999999999, total_cost = 2013.9974999999999, pathkeys = 0x0} (gdb) p *$np->outerjoinpath $110 = {type = T_HashPath, pathtype = T_HashJoin, parent = 0x2f54050, pathtarget = 0x2fcbf88, param_info = 0x0, parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 100, startup_cost = 3.25, total_cost = 196.75, pathkeys = 0x0}到此,关于“***中使用动态规划算法构造连接路径的实现函数是哪个”的学习就结束了,希望能够解决大家的疑惑。理论与实践的搭配能更好的帮助大家学习,快去试试吧!若想继续学习更多相关知识,请继续关注网站,小编会继续努力为大家带来更多实用的文章!
版权声明:本文内容由网络用户投稿,版权归原作者所有,本站不拥有其著作权,亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容,请联系我们jiasou666@gmail.com 处理,核实后本网站将在24小时内删除侵权内容。