pengkingli 2019-07-01
接着 https://mengkang.net/1328.html 的案例,我们继续磕。
上一篇 GDB 调试 Mysql 实战(三)优先队列排序算法探究(上) 分析了实验3中的row_size
为什么是24。其他实验的row_size
都是36,扫描行数也不符合预期。这篇就来探究下。
以实验1为例来分析
select `aid`,sum(`pv`) as num from article_rank force index(idx_day_aid_pv) where `day`>20190115 group by aid order by num desc LIMIT 10;
optimizer_trace.join_execution.steps
的结果如下
{ "join_execution": { "select#": 1, "steps": [ { "creating_tmp_table": { "tmp_table_info": { "table": "intermediate_tmp_table", "row_length": 20, "key_length": 4, "unique_constraint": false, "location": "memory (heap)", "row_limit_estimate": 838860 } } }, { "converting_tmp_table_to_ondisk": { "cause": "memory_table_size_exceeded", "tmp_table_info": { "table": "intermediate_tmp_table", "row_length": 20, "key_length": 4, "unique_constraint": false, "location": "disk (InnoDB)", "record_format": "fixed" } } }, { "filesort_information": [ { "direction": "desc", "table": "intermediate_tmp_table", "field": "num" } ], "filesort_priority_queue_optimization": { "limit": 10, "rows_estimate": 1057, "row_size": 36, "memory_available": 262144, "chosen": true }, "filesort_execution": [ ], "filesort_summary": { "rows": 11, "examined_rows": 649091, "number_of_tmp_files": 0, "sort_buffer_size": 488, "sort_mode": "<sort_key, additional_fields>" } } ] } }
(gdb) b Sort_param::init_for_filesort Breakpoint 1 at 0xf1a89f: file /root/newdb/mysql-server/sql/filesort.cc, line 107.
(gdb) b Filesort::get_addon_fields Breakpoint 2 at 0xf21231: file /root/newdb/mysql-server/sql/filesort.cc, line 2459. (gdb) b /root/newdb/mysql-server/sql/filesort.cc:2496 Breakpoint 3 at 0xf212f9: file /root/newdb/mysql-server/sql/filesort.cc, line 2496. (gdb) b /root/newdb/mysql-server/sql/filesort.cc:2523 Breakpoint 4 at 0xf2145f: file /root/newdb/mysql-server/sql/filesort.cc, line 2523.
排序字段还是实验3一样是16字节,后面20字节则是两个字段相加20字节+ (null_fields + 7) / 8
一个可为空的字段,所以最后是36了。
(gdb) b /root/newdb/mysql-server/sql/filesort.cc:320 Breakpoint 5 at 0xf1b1d9: file /root/newdb/mysql-server/sql/filesort.cc, line 320. ... Breakpoint 5, filesort (thd=0x7f0214000d80, filesort=0x7f021401f668, sort_positions=false, examined_rows=0x7f022804d050, found_rows=0x7f022804d048, returned_rows=0x7f022804d040) at /root/newdb/mysql-server/sql/filesort.cc:320 320 num_rows= table->file->estimate_rows_upper_bound(); (gdb) s ha_innobase::estimate_rows_upper_bound (this=0x7f0214022b50) at /root/newdb/mysql-server/storage/innobase/handler/ha_innodb.cc:13655 ha_innobase::estimate_rows_upper_bound (this=0x7f0214022b50) at /root/newdb/mysql-server/storage/innobase/handler/ha_innodb.cc:13655 warning: Source file is more recent than executable. 13655 DBUG_ENTER("estimate_rows_upper_bound"); (gdb) n 13661 update_thd(ha_thd()); (gdb) n 13663 TrxInInnoDB trx_in_innodb(m_prebuilt->trx); (gdb) n 13665 m_prebuilt->trx->op_info = "calculating upper bound for table rows"; (gdb) n 13667 index = dict_table_get_first_index(m_prebuilt->table); (gdb) n 13669 ulint stat_n_leaf_pages = index->stat_n_leaf_pages; (gdb) p stat_n_leaf_pages $19 = 139646902217632 (gdb) n 13671 ut_a(stat_n_leaf_pages > 0); (gdb) p UNIV_PAGE_SIZE No symbol "UNIV_PAGE_SIZE" in current context. (gdb) n 13674 ((ulonglong) stat_n_leaf_pages) * UNIV_PAGE_SIZE; (gdb) n 13681 estimate = 2 * local_data_file_length (gdb) p local_data_file_length $20 = 16384 (gdb) p stat_n_leaf_pages $21 = 1 (gdb) n 13682 / dict_index_calc_min_rec_len(index); (gdb) n 13684 m_prebuilt->trx->op_info = ""; (gdb) p estimate $22 = 1057 (gdb) p dict_index_calc_min_rec_len(index) $23 = 31
也就是说local_data_file_length
是16字节,为当前系统一个内存页大小。dict_index_calc_min_rec_len
注释中写道Calculates the minimum record length in an index.
上面gdb调试记录(文字非截图)中dict_index_calc_min_rec_len(index)
的值为31。
ut_a(stat_n_leaf_pages > 0); local_data_file_length = ((ulonglong) stat_n_leaf_pages) * UNIV_PAGE_SIZE; /* Calculate a minimum length for a clustered index record and from that an upper bound for the number of rows. Since we only calculate new statistics in row0mysql.cc when a table has grown by a threshold factor, we must add a safety factor 2 in front of the formula below. */ estimate = 2 * local_data_file_length / dict_index_calc_min_rec_len(index);
(2*16*1024)/31 = 1057
,那么为什么dict_index_calc_min_rec_len
是31呢?
继续查看源码发现 31 是这么计算出来的,就算知道了31,但是我也还是木有弄懂,为什么扫描行数是 (2*页内存大小)/索引最小行记录长度
要知道时间复杂度只是描述一个增长趋势,复杂度为O的排序算法执行时间不一定比复杂度为O长,因为在计算O时省略了系数、常数、低阶。实际上,在对小规模数据进行排序时,n2的值实际比 knlogn+c还要小。