PostgreSQL物理优化中的create_index_paths->choose_bitmap_and函数分析

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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_andBreakpoint 1 at 0x74e8c2: file indxpath.c, line 1372.(gdb) cContinuing.Breakpoint 1, choose_bitmap_and (root=0x1666638, rel=0x1666a48, paths=0x166fdf0) at indxpath.c:13721372    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) n1471        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) n1480    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) n1495      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) n1498      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) n1531        newcost = bitmap_and_cost_est(root, rel, paths);(gdb) 1532        if (newcost < costsofar)(gdb) p newcost$30 = 88.375456720095343(gdb) n1535          costsofar = newcost;(gdb) n1537                      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) n1498      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) n1563    if (list_length(bestpaths) == 1)(gdb) 1565    return (Path *) create_bitmap_and_path(root, rel, bestpaths);(gdb) 1566  }

DONE!

(gdb) ncreate_index_paths (root=0x1666638, rel=0x1666a48) at indxpath.c:337337     bpath = create_bitmap_heap_path(root, rel, bitmapqual,

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