这篇文章主要讲解了“PostgreSQL物理优化中的create_index_paths->choose_bitmap_and函数分析”,文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习“PostgreSQL物理优化中的create_index_paths->choose_bitmap_and函数分析”吧!
该函数执行Bitmap AND操作后创建位图索引扫描访问路径(BitmapAndPath)节点。
下面是BitmapAnd访问路径的样例:
testdb=# explain verbose select t1.* testdb-# from t_dwxx t1 testdb-# where (dwbh > '10000' and dwbh < '15000') AND (dwdz between 'DWDZ10000' and 'DWDZ15000'); QUERY PLAN ---------------------------------------------------------------------------------------------- Bitmap Heap Scan on public.t_dwxx t1 (cost=32.33..88.38 rows=33 width=20) Output: dwmc, dwbh, dwdz Recheck Cond: (((t1.dwbh)::text > '10000'::text) AND ((t1.dwbh)::text < '15000'::text) AND ((t1.dwdz)::text >= 'DWDZ10000' ::text) AND ((t1.dwdz)::text <= 'DWDZ15000'::text)) -> BitmapAnd (cost=32.33..32.33 rows=33 width=0) -->BitmapAnd -> Bitmap Index Scan on t_dwxx_pkey (cost=0.00..13.86 rows=557 width=0) Index Cond: (((t1.dwbh)::text > '10000'::text) AND ((t1.dwbh)::text < '15000'::text)) -> Bitmap Index Scan on idx_dwxx_dwdz (cost=0.00..18.21 rows=592 width=0) Index Cond: (((t1.dwdz)::text >= 'DWDZ10000'::text) AND ((t1.dwdz)::text <= 'DWDZ15000'::text)) (8 rows)
一、数据结构
Cost相关
注意:实际使用的参数值通过系统配置文件定义,而不是这里的常量定义!
typedef double Cost; /* execution cost (in page-access units) */ /* defaults for costsize.c's Cost parameters */ /* NB: cost-estimation code should use the variables, not these constants! */ /* 注意:实际值通过系统配置文件定义,而不是这里的常量定义! */ /* If you change these, update backend/utils/misc/postgresql.sample.conf */ #define DEFAULT_SEQ_PAGE_COST 1.0 //顺序扫描page的成本 #define DEFAULT_RANDOM_PAGE_COST 4.0 //随机扫描page的成本 #define DEFAULT_CPU_TUPLE_COST 0.01 //处理一个元组的CPU成本 #define DEFAULT_CPU_INDEX_TUPLE_COST 0.005 //处理一个索引元组的CPU成本 #define DEFAULT_CPU_OPERATOR_COST 0.0025 //执行一次操作或函数的CPU成本 #define DEFAULT_PARALLEL_TUPLE_COST 0.1 //并行执行,从一个worker传输一个元组到另一个worker的成本 #define DEFAULT_PARALLEL_SETUP_COST 1000.0 //构建并行执行环境的成本 #define DEFAULT_EFFECTIVE_CACHE_SIZE 524288 /*先前已有介绍, measured in pages */ double seq_page_cost = DEFAULT_SEQ_PAGE_COST; double random_page_cost = DEFAULT_RANDOM_PAGE_COST; double cpu_tuple_cost = DEFAULT_CPU_TUPLE_COST; double cpu_index_tuple_cost = DEFAULT_CPU_INDEX_TUPLE_COST; double cpu_operator_cost = DEFAULT_CPU_OPERATOR_COST; double parallel_tuple_cost = DEFAULT_PARALLEL_TUPLE_COST; double parallel_setup_cost = DEFAULT_PARALLEL_SETUP_COST; int effective_cache_size = DEFAULT_EFFECTIVE_CACHE_SIZE; Cost disable_cost = 1.0e10;//1后面10个0,通过设置一个巨大的成本,让优化器自动放弃此路径 int max_parallel_workers_per_gather = 2;//每次gather使用的worker数
PathClauseUsage
/* Per-path data used within choose_bitmap_and() */ typedef struct { Path *path; /* 访问路径链表,IndexPath, BitmapAndPath, or BitmapOrPath */ List *quals; /* 限制条件子句链表,the WHERE clauses it uses */ List *preds; /* 部分索引谓词链表,predicates of its partial index(es) */ Bitmapset *clauseids; /* 位图集合,quals+preds represented as a bitmapset */ } PathClauseUsage;
二、源码解读
choose_bitmap_and函数
create_index_paths->choose_bitmap_and函数,该函数给定非空的位图访问路径链表,执行AND操作后合并到一条路径中,最终得到位图索引扫描访问路径节点.
/* * choose_bitmap_and * Given a nonempty list of bitmap paths, AND them into one path. * 给定非空的位图访问路径链表,执行AND操作后合并到一条路径中 * * This is a nontrivial decision since we can legally use any subset of the * given path set. We want to choose a good tradeoff between selectivity * and cost of computing the bitmap. * 这是一个非常重要的策略,因为这样可以合法地使用给定路径集的任何子集。 * * The result is either a single one of the inputs, or a BitmapAndPath * combining multiple inputs. * 输出结果要么是输出的其中之一,要么是融合多个输入之后的BitmapAndPath */ static Path * choose_bitmap_and(PlannerInfo *root, RelOptInfo *rel, List *paths) { int npaths = list_length(paths); PathClauseUsage **pathinfoarray; PathClauseUsage *pathinfo; List *clauselist; List *bestpaths = NIL; Cost bestcost = 0; int i, j; ListCell *l; Assert(npaths > 0); /* else caller error */ if (npaths == 1) return (Path *) linitial(paths); /* easy case */ /* * In theory we should consider every nonempty subset of the given paths. * In practice that seems like overkill, given the crude nature of the * estimates, not to mention the possible effects of higher-level AND and * OR clauses. Moreover, it's completely impractical if there are a large * number of paths, since the work would grow as O(2^N). * 理论上,我们应该考虑给定路径的所有非空子集。在实践中, * 考虑到估算的不确定性和成本,以及更高级别的AND和OR约束可能产生的影响,这样的做法并不合适. * 此外,它并不切合实际,如果有大量的路径,这项工作的复杂度会是指数级的O(2 ^ N)。 * * As a heuristic, we first check for paths using exactly the same sets of * WHERE clauses + index predicate conditions, and reject all but the * cheapest-to-scan in any such group. This primarily gets rid of indexes * that include the interesting columns but also irrelevant columns. (In * situations where the DBA has gone overboard on creating variant * indexes, this can make for a very large reduction in the number of * paths considered further.) * 作为一种启发式方法,首先使用完全相同的WHERE子句+索引谓词条件集检查路径, * 并去掉这类条件组中除成本最低之外的所有路径。 * 这主要是去掉了包含interesting列和不相关列的索引。 * (在DBA过度创建索引的情况下,这会大大减少进一步考虑的路径数量。) * * We then sort the surviving paths with the cheapest-to-scan first, and * for each path, consider using that path alone as the basis for a bitmap * scan. Then we consider bitmap AND scans formed from that path plus * each subsequent (higher-cost) path, adding on a subsequent path if it * results in a reduction in the estimated total scan cost. This means we * consider about O(N^2) rather than O(2^N) path combinations, which is * quite tolerable, especially given than N is usually reasonably small * because of the prefiltering step. The cheapest of these is returned. * 然后,我们首先使用成本最低的扫描路径对现存的路径进行排序, * 对于每个路径,考虑单独使用该路径作为位图扫描的基础。 * 然后我们考虑位图和从该路径形成的扫描加上每个后续的(更高成本的)路径, * 如果后续路径导致估算的总扫描成本减少,那么就添加一个后续路径。 * 这意味着我们只需要处理O(N ^ 2),而不是O(2 ^ N)个路径组合, * 这样的成本完全可以接受,特别是N通常相当小时。函数返回成本最低的路径。 * * We will only consider AND combinations in which no two indexes use the * same WHERE clause. This is a bit of a kluge: it's needed because * costsize.c and clausesel.c aren't very smart about redundant clauses. * They will usually double-count the redundant clauses, producing a * too-small selectivity that makes a redundant AND step look like it * reduces the total cost. Perhaps someday that code will be smarter and * we can remove this limitation. (But note that this also defends * against flat-out duplicate input paths, which can happen because * match_join_clauses_to_index will find the same OR join clauses that * extract_restriction_or_clauses has pulled OR restriction clauses out * of.) * 我们将只考虑没有两个索引同时使用相同的WHERE子句的AND组合。 * 这是一个有点蹩脚的做法:之所以这样是因为cost.c和clausesel.c未能足够聪明的处理多余的子句。 * 它们通常会重复计算冗余子句,从而产生很小的选择性,使冗余子句看起来像是减少了总成本。 * 也许有一天,代码会变得更聪明,我们可以消除这个限制。 * (但是要注意,这也可以防止完全重复的输入路径, * 因为match_join_clauses_to_index会找到相同的OR连接子句,而这些子句 * 已通过extract_restriction_or_clauses函数提升到外面去了.) * * For the same reason, we reject AND combinations in which an index * predicate clause duplicates another clause. Here we find it necessary * to be even stricter: we'll reject a partial index if any of its * predicate clauses are implied by the set of WHERE clauses and predicate * clauses used so far. This covers cases such as a condition "x = 42" * used with a plain index, followed by a clauseless scan of a partial * index "WHERE x >= 40 AND x < 50". The partial index has been accepted * only because "x = 42" was present, and so allowing it would partially * double-count selectivity. (We could use predicate_implied_by on * regular qual clauses too, to have a more intelligent, but much more * expensive, check for redundancy --- but in most cases simple equality * seems to suffice.) * 出于同样的原因,我们不会组合索引谓词子句与另一个重复的子句。 * 在这里,有必要更加严格 : 如果部分索引的任何谓词子句 * 隐含在WHERE子句中,则不能使用此索引。 * 这里包括了形如使用普通索引的“x = 42”和使用部分索引“x >= 40和x < 50”的情况。 * 部分索引被接受,是因为存在“x = 42”,因此允许它部分重复计数选择性。 * (我们也可以在普通的qual子句上使用predicate_implied_by函数, * 这样就可以更智能但更昂贵地检查冗余——但在大多数情况下,简单的等式似乎就足够了。) */ /* * Extract clause usage info and detect any paths that use exactly the * same set of clauses; keep only the cheapest-to-scan of any such groups. * The surviving paths are put into an array for qsort'ing. * 提取子句使用信息并检测使用完全相同子句集的所有路径; * 只保留这类路径中成本最低的,这些路径被放入一个数组中进行qsort'ing */ pathinfoarray = (PathClauseUsage **) palloc(npaths * sizeof(PathClauseUsage *));//数组 clauselist = NIL; npaths = 0; foreach(l, paths)//遍历paths { Path *ipath = (Path *) lfirst(l); pathinfo = classify_index_clause_usage(ipath, &clauselist);//归类路径信息 for (i = 0; i < npaths; i++) { if (bms_equal(pathinfo->clauseids, pathinfoarray[i]->clauseids)) break;//只要发现子句集一样,就继续执行 } if (i < npaths)//发现相同的 { /* duplicate clauseids, keep the cheaper one */ //相同的约束条件,只保留成本最低的 Cost ncost; Cost ocost; Selectivity nselec; Selectivity oselec; cost_bitmap_tree_node(pathinfo->path, &ncost, &nselec);//计算成本 cost_bitmap_tree_node(pathinfoarray[i]->path, &ocost, &oselec); if (ncost < ocost) pathinfoarray[i] = pathinfo; } else//没有发现条件一样的,添加到数组中 { /* not duplicate clauseids, add to array */ pathinfoarray[npaths++] = pathinfo; } } /* If only one surviving path, we're done */ if (npaths == 1)//结果只有一条,则返回之 return pathinfoarray[0]->path; /* Sort the surviving paths by index access cost */ qsort(pathinfoarray, npaths, sizeof(PathClauseUsage *), path_usage_comparator);//以索引访问成本排序现存路径 /* * For each surviving index, consider it as an "AND group leader", and see * whether adding on any of the later indexes results in an AND path with * cheaper total cost than before. Then take the cheapest AND group. * 对于现存的索引,把它视为"AND group leader", * 并查看是否添加了以后的索引后,会得到一个总成本比以前更低的AND路径。 * 选择成本最低的AND组. * */ for (i = 0; i < npaths; i++)//遍历这些路径 { Cost costsofar; List *qualsofar; Bitmapset *clauseidsofar; ListCell *lastcell; pathinfo = pathinfoarray[i];//PathClauseUsage结构体 paths = list_make1(pathinfo->path);//路径链表 costsofar = bitmap_scan_cost_est(root, rel, pathinfo->path);//当前的成本 qualsofar = list_concat(list_copy(pathinfo->quals), list_copy(pathinfo->preds)); clauseidsofar = bms_copy(pathinfo->clauseids); lastcell = list_head(paths); /* 用于快速删除,for quick deletions */ for (j = i + 1; j < npaths; j++)//扫描后续的路径 { Cost newcost; pathinfo = pathinfoarray[j]; /* Check for redundancy */ if (bms_overlap(pathinfo->clauseids, clauseidsofar)) continue; /* 多余的路径,consider it redundant */ if (pathinfo->preds)//部分索引? { bool redundant = false; /* we check each predicate clause separately */ //单独检查每一个谓词 foreach(l, pathinfo->preds) { Node *np = (Node *) lfirst(l); if (predicate_implied_by(list_make1(np), qualsofar, false)) { redundant = true; break; /* out of inner foreach loop */ } } if (redundant) continue; } /* tentatively add new path to paths, so we can estimate cost */ //尝试在路径中添加新路径,这样我们就可以估算成本 paths = lappend(paths, pathinfo->path); newcost = bitmap_and_cost_est(root, rel, paths);//估算成本 if (newcost < costsofar)//新成本更低 { /* keep new path in paths, update subsidiary variables */ costsofar = newcost; qualsofar = list_concat(qualsofar, list_copy(pathinfo->quals));//添加此条件 qualsofar = list_concat(qualsofar, list_copy(pathinfo->preds));//添加此谓词 clauseidsofar = bms_add_members(clauseidsofar, pathinfo->clauseids);//添加此子句ID lastcell = lnext(lastcell); } else { /* reject new path, remove it from paths list */ paths = list_delete_cell(paths, lnext(lastcell), lastcell);//去掉新路径 } Assert(lnext(lastcell) == NULL); } /* Keep the cheapest AND-group (or singleton) */ if (i == 0 || costsofar < bestcost)//单条路径或者取得最小的成本 { bestpaths = paths; bestcost = costsofar; } /* some easy cleanup (we don't try real hard though) */ list_free(qualsofar); } if (list_length(bestpaths) == 1) return (Path *) linitial(bestpaths); /* 无需AND路径,no need for AND */ return (Path *) create_bitmap_and_path(root, rel, bestpaths);//生成BitmapAndPath } //-------------------------------------------------------------------------- bitmap_scan_cost_est /* * Estimate the cost of actually executing a bitmap scan with a single * index path (no BitmapAnd, at least not at this level; but it could be * a BitmapOr). */ static Cost bitmap_scan_cost_est(PlannerInfo *root, RelOptInfo *rel, Path *ipath) { BitmapHeapPath bpath; Relids required_outer; /* Identify required outer rels, in case it's a parameterized scan */ required_outer = get_bitmap_tree_required_outer(ipath); /* Set up a dummy BitmapHeapPath */ bpath.path.type = T_BitmapHeapPath; bpath.path.pathtype = T_BitmapHeapScan; bpath.path.parent = rel; bpath.path.pathtarget = rel->reltarget; bpath.path.param_info = get_baserel_parampathinfo(root, rel, required_outer); bpath.path.pathkeys = NIL; bpath.bitmapqual = ipath; /* * Check the cost of temporary path without considering parallelism. * Parallel bitmap heap path will be considered at later stage. */ bpath.path.parallel_workers = 0; cost_bitmap_heap_scan(&bpath.path, root, rel, bpath.path.param_info, ipath, get_loop_count(root, rel->relid, required_outer));//BitmapHeapPath计算成本 return bpath.path.total_cost; } //-------------------------------------------------------------------------- bitmap_and_cost_est /* * Estimate the cost of actually executing a BitmapAnd scan with the given * inputs. * 给定输入,估算实际执行BitmapAnd扫描的实际成本 */ static Cost bitmap_and_cost_est(PlannerInfo *root, RelOptInfo *rel, List *paths) { BitmapAndPath apath; BitmapHeapPath bpath; Relids required_outer; /* Set up a dummy BitmapAndPath */ apath.path.type = T_BitmapAndPath; apath.path.pathtype = T_BitmapAnd; apath.path.parent = rel; apath.path.pathtarget = rel->reltarget; apath.path.param_info = NULL; /* not used in bitmap trees */ apath.path.pathkeys = NIL; apath.bitmapquals = paths; cost_bitmap_and_node(&apath, root); /* Identify required outer rels, in case it's a parameterized scan */ required_outer = get_bitmap_tree_required_outer((Path *) &apath); /* Set up a dummy BitmapHeapPath */ bpath.path.type = T_BitmapHeapPath; bpath.path.pathtype = T_BitmapHeapScan; bpath.path.parent = rel; bpath.path.pathtarget = rel->reltarget; bpath.path.param_info = get_baserel_parampathinfo(root, rel, required_outer); bpath.path.pathkeys = NIL; bpath.bitmapqual = (Path *) &apath; /* * Check the cost of temporary path without considering parallelism. * Parallel bitmap heap path will be considered at later stage. */ bpath.path.parallel_workers = 0; /* Now we can do cost_bitmap_heap_scan */ cost_bitmap_heap_scan(&bpath.path, root, rel, bpath.path.param_info, (Path *) &apath, get_loop_count(root, rel->relid, required_outer));//BitmapHeapPath计算成本 return bpath.path.total_cost; } //-------------------------------------------------------------------------- create_bitmap_and_path /* * create_bitmap_and_path * Creates a path node representing a BitmapAnd. */ BitmapAndPath * create_bitmap_and_path(PlannerInfo *root, RelOptInfo *rel, List *bitmapquals) { BitmapAndPath *pathnode = makeNode(BitmapAndPath); pathnode->path.pathtype = T_BitmapAnd; pathnode->path.parent = rel; pathnode->path.pathtarget = rel->reltarget; pathnode->path.param_info = NULL; /* not used in bitmap trees */ /* * Currently, a BitmapHeapPath, BitmapAndPath, or BitmapOrPath will be * parallel-safe if and only if rel->consider_parallel is set. So, we can * set the flag for this path based only on the relation-level flag, * without actually iterating over the list of children. */ pathnode->path.parallel_aware = false; pathnode->path.parallel_safe = rel->consider_parallel; pathnode->path.parallel_workers = 0; pathnode->path.pathkeys = NIL; /* always unordered */ pathnode->bitmapquals = bitmapquals; /* this sets bitmapselectivity as well as the regular cost fields: */ cost_bitmap_and_node(pathnode, root);//计算成本 return pathnode; } //----------------------------------------------------- cost_bitmap_and_node /* * cost_bitmap_and_node * Estimate the cost of a BitmapAnd node * 估算BitmapAnd节点成本 * * Note that this considers only the costs of index scanning and bitmap * creation, not the eventual heap access. In that sense the object isn't * truly a Path, but it has enough path-like properties (costs in particular) * to warrant treating it as one. We don't bother to set the path rows field, * however. */ void cost_bitmap_and_node(BitmapAndPath *path, PlannerInfo *root) { Cost totalCost; Selectivity selec; ListCell *l; /* * We estimate AND selectivity on the assumption that the inputs are * independent. This is probably often wrong, but we don't have the info * to do better. * * The runtime cost of the BitmapAnd itself is estimated at 100x * cpu_operator_cost for each tbm_intersect needed. Probably too small, * definitely too simplistic? */ totalCost = 0.0; selec = 1.0; foreach(l, path->bitmapquals) { Path *subpath = (Path *) lfirst(l); Cost subCost; Selectivity subselec; cost_bitmap_tree_node(subpath, &subCost, &subselec); selec *= subselec; totalCost += subCost; if (l != list_head(path->bitmapquals)) totalCost += 100.0 * cpu_operator_cost; } path->bitmapselectivity = selec; path->path.rows = 0; /* per above, not used */ path->path.startup_cost = totalCost; path->path.total_cost = totalCost; }
三、跟踪分析
测试脚本如下
select t1.* from t_dwxx t1 where (dwbh > '10000' and dwbh < '15000') AND (dwdz between 'DWDZ10000' and 'DWDZ15000');
启动gdb跟踪
(gdb) b choose_bitmap_and Breakpoint 1 at 0x74e8c2: file indxpath.c, line 1372. (gdb) c Continuing. Breakpoint 1, choose_bitmap_and (root=0x1666638, rel=0x1666a48, paths=0x166fdf0) at indxpath.c:1372 1372 int npaths = list_length(paths);
输入参数
(gdb) p *paths $1 = {type = T_List, length = 2, head = 0x166fe20, tail = 0x16706b8} (gdb) p *(Node *)paths->head->data.ptr_value $2 = {type = T_IndexPath} (gdb) p *(Node *)paths->head->next->data.ptr_value $3 = {type = T_IndexPath} (gdb) set $p1=(IndexPath *)paths->head->data.ptr_value (gdb) set $p2=(IndexPath *)paths->head->next->data.ptr_value (gdb) p *$p1 $4 = {path = {type = T_IndexPath, pathtype = T_IndexScan, parent = 0x1666a48, pathtarget = 0x166d988, param_info = 0x0, parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 33, startup_cost = 0.28500000000000003, total_cost = 116.20657683302848, pathkeys = 0x0}, indexinfo = 0x166e420, indexclauses = 0x166f528, indexquals = 0x166f730, indexqualcols = 0x166f780, indexorderbys = 0x0, indexorderbycols = 0x0, indexscandir = ForwardScanDirection, indextotalcost = 18.205000000000002, indexselectivity = 0.059246954595791879} (gdb) p *$p2 $5 = {path = {type = T_IndexPath, pathtype = T_IndexScan, parent = 0x1666a48, pathtarget = 0x166d988, param_info = 0x0, parallel_aware = false, parallel_safe = true, parallel_workers = 0, rows = 33, startup_cost = 0.28500000000000003, total_cost = 111.33157683302848, pathkeys = 0x0}, indexinfo = 0x1666c58, indexclauses = 0x166fed0, indexquals = 0x166ffc8, indexqualcols = 0x1670018, indexorderbys = 0x0, indexorderbycols = 0x0, indexscandir = ForwardScanDirection, indextotalcost = 13.855, indexselectivity = 0.055688888888888899}
paths中的第1个元素对应(dwbh > '10000' and dwbh < '15000') ,第2个元素对应(dwdz between 'DWDZ10000' and 'DWDZ15000')
(gdb) set $ri1=(RestrictInfo *)$p1->indexclauses->head->data.ptr_value (gdb) set $tmp=(RelabelType *)((OpExpr *)$ri1->clause)->args->head->data.ptr_value (gdb) p *(Var *)$tmp->arg $17 = {xpr = {type = T_Var}, varno = 1, varattno = 3, vartype = 1043, vartypmod = 104, varcollid = 100, varlevelsup = 0, varnoold = 1, varoattno = 3, location = 76} (gdb) p *(Node *)((OpExpr *)$ri1->clause)->args->head->next->data.ptr_value $18 = {type = T_Const} (gdb) p *(Const *)((OpExpr *)$ri1->clause)->args->head->next->data.ptr_value $19 = {xpr = {type = T_Const}, consttype = 25, consttypmod = -1, constcollid = 100, constlen = -1, constvalue = 23636608, constisnull = false, constbyval = false, location = 89}
开始遍历paths,提取子句条件并检测是否使用完全相同子句集的所有路径,只保留这些路径中成本最低的,这些路径被放入一个数组中进行qsort.
... (gdb) 1444 npaths = 0; (gdb) 1445 foreach(l, paths) (gdb)
收集信息到PathClauseUsage数组中
... (gdb) n 1471 pathinfoarray[npaths++] = pathinfo; (gdb) 1445 foreach(l, paths) (gdb) 1476 if (npaths == 1) (gdb) p npaths $26 = 2 (gdb)
按成本排序
(gdb) n 1480 qsort(pathinfoarray, npaths, sizeof(PathClauseUsage *),
遍历路径,找到成本最低的AND group
1488 for (i = 0; i < npaths; i++) (gdb) n 1495 pathinfo = pathinfoarray[i]; (gdb) 1496 paths = list_make1(pathinfo->path); (gdb) 1497 costsofar = bitmap_scan_cost_est(root, rel, pathinfo->path); (gdb) 1499 list_copy(pathinfo->preds));
获取当前的成本,设置当前的条件子句
(gdb) p costsofar $27 = 89.003250000000008 (gdb) n 1498 qualsofar = list_concat(list_copy(pathinfo->quals),
执行AND操作(路径叠加),成本更低,调整当前成本和相关变量
(gdb) n 1531 newcost = bitmap_and_cost_est(root, rel, paths); (gdb) 1532 if (newcost < costsofar) (gdb) p newcost $30 = 88.375456720095343 (gdb) n 1535 costsofar = newcost; (gdb) n 1537 list_copy(pathinfo->quals)); (gdb) 1536 qualsofar = list_concat(qualsofar, (gdb) 1539 list_copy(pathinfo->preds));
处理下一个AND条件,单个AND条件比上一个条件成本高,保留原来的
1488 for (i = 0; i < npaths; i++) (gdb) 1495 pathinfo = pathinfoarray[i]; (gdb) 1496 paths = list_make1(pathinfo->path); (gdb) 1497 costsofar = bitmap_scan_cost_est(root, rel, pathinfo->path); (gdb) 1499 list_copy(pathinfo->preds)); (gdb) p costsofar $34 = 94.053250000000006 (gdb) n 1498 qualsofar = list_concat(list_copy(pathinfo->quals), (gdb) 1500 clauseidsofar = bms_copy(pathinfo->clauseids); (gdb) 1501 lastcell = list_head(paths); /* for quick deletions */ (gdb) 1503 for (j = i + 1; j < npaths; j++) (gdb) 1553 if (i == 0 || costsofar < bestcost) (gdb) p i $35 = 1 (gdb) p costsofar $36 = 94.053250000000006 (gdb) p bestcost $37 = 88.375456720095343 (gdb)
构建BitmapAndPath,返回
(gdb) n 1563 if (list_length(bestpaths) == 1) (gdb) 1565 return (Path *) create_bitmap_and_path(root, rel, bestpaths); (gdb) 1566 }
DONE!
(gdb) n create_index_paths (root=0x1666638, rel=0x1666a48) at indxpath.c:337 337 bpath = create_bitmap_heap_path(root, rel, bitmapqual,
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原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/205207.html