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！
成都创新互联公司-专业网站定制、快速模板网站建设、高性价比龙陵网站开发、企业建站全套包干低至880元,成熟完善的模板库,直接使用。一站式龙陵网站制作公司更省心,省钱,快速模板网站建设找我们，业务覆盖龙陵地区。费用合理售后完善，十多年实体公司更值得信赖。
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,
感谢各位的阅读，以上就是“PostgreSQL物理优化中的create_index_paths->choose_bitmap_and函数分析”的内容了，经过本文的学习后，相信大家对PostgreSQL物理优化中的create_index_paths->choose_bitmap_and函数分析这一问题有了更深刻的体会，具体使用情况还需要大家实践验证。这里是创新互联，小编将为大家推送更多相关知识点的文章，欢迎关注！
网页名称：PostgreSQL物理优化中的create_index_paths->choose_bitmap_and函数分析
网站网址：
http://cdkjz.cn/article/jsjhep.html
