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物理读之LRU（最近最少被使用）的深入解析
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http://blog.csdn.net/guoyjoe/article/details/38264883
一组LRU链表包括LRU主链，LRU辅助链，LRUW主链，LRUW辅助链，称为一个WorkSet(工作组)如下图：
sys@ZMDB> select CNUM_SET,CNUM_REPL,ANUM_REPL,CNUM_WRITE,ANUM_WRITE from x$kcbwds where CNUM_SET>0;
CNUM_SET CNUM_REPL ANUM_REPL CNUM_WRITE ANUM_WRITE
---------- ---------- ---------- ---------- ----------
15221 15221 3796 0 0
15221 15221 3783 0 0
CNUM_SET:工作组总的buffer总数量
CNUM_REPL:工作组中LRU的buffer总数量（主LRU+辅LRU）
ANUM_REPL:工作组中辅LRU总BUFFER的数量
通过隐含参数查到BUFFER的总的个数是30442，正好与上面的CNUM_SET=15221+15221
sys@ZMDB> @?/rdbms/admin/show_para
Enter value for p: _db_block_buffers
old 12: AND upper(i.ksppinm) LIKE upper(‘%&p%‘)
new 12: AND upper(i.ksppinm) LIKE upper(‘%_db_block_buffers%‘)
P_NAME P_DESCRIPTION P_VALUE ISDEFAULT ISMODIFIED ISADJ
---------------------------------------- -------------------------------------------------- ------------------------------ --------- ---------- -----
_db_block_buffers Number of database blocks cached in memory: hidden 30442 TRUE FALSE FALSE
Parameter
我们用以下语句查下数据库中buffer所在LRU的状态
sys@ZMDB> select lru_flag,count(*) from x$bh group by lru_flag;
LRU_FLAG COUNT(*)
---------- ----------
6 208
2 10
4 7122
8 15199
0 7646
我们对LRU_FLAG=6，2，4，8，0等做出解释，举个例子，对于6是什么含义呢？
首先要在x$bh中找到lru_flag=6的任意的一个BUFFER
sys@ZMDB> select LRU_FLAG,LOWER(BA)from x$bh where lru_flag=6 and rownum=1;
LRU_FLAG LOWER(BA)
---------- ----------------
6 0000000081dae000
DUMP buffer_cache中BH信息，如下命令：
sys@ZMDB> alter session set events‘immediate trace name buffers level 1‘; Session altered. ys@ZMDB> col value for a85 sys@ZMDB> select * from v$diag_info where name=‘Default Trace File‘; INST_ID NAME VALUE ---------- -------------------------------------------------- ------------------------------------------------------------------------------------- 1 Default Trace File /u01/app/oracle/diag/rdbms/zmdb/zmdb/trace/zmdb_ora_13235.trc
通过BA=81dae000搜索trace文件，
/u01/app/oracle/diag/rdbms/zmdb/zmdb/trace/zmdb_ora_13235.trc
得到如下内容：
BH (0x81fe7e38) file#: 1 rdba: 0x0040ace1 (1/44257) class: 1 ba: 0x81dae000 set: 6 pool: 3 bsz: 8192 bsi: 0 sflg: 1 pwc: 0,25 dbwrid: 0 obj: 421 objn: 423 tsn: 0 afn: 1 hint: f hash: [0x9ef9d710,0x853f8da8] lru: [0x81fe7df0,0x81fe8050] lru-flags: moved_to_tail on_auxiliary_list ckptq: [NULL] fileq: [NULL] objq: [NULL] objaq: [NULL] st: CR md: NULL fpin: ‘kdswh06: kdscgr‘ tch: 1 cr: [scn: 0x0.80350f4d],[xid: 0x0.0.0],[uba: 0x0.0.0],[cls: 0x0.80350f4d],[sfl: 0x0],[lc: 0x0.8034c532] flags: block_written_once redo_since_read
LRU_FLAG=6的意思是lru-flags:moved_to_tail on_auxiliary_list，就是向LRU的辅助链表的尾部移动，这有可能是SMON从LRU的主链表上的非脏块、TCH<=1并且状态是非PIN的BUFFER被挂接到LRU辅助链表的尾部。
根据以上的方法同理可以解释出LRU_FLAG的含义：
LRU_FLAG
0==>LRU-主链冷端的头部，这个比较特殊他在DUMP没有显示LRU_FLAG
2==>LRU-主链冷端的尾部，lru-flags:moved_to_tail
4==>LRU-辅助链,lru-flags:on_auxiliary_list
6==>LRU-辅助链的尾部，lru-flags:moved_to_tail on_auxiliary_list
8==>LUR-主链热端，lru-flags:hot_buffer
当发生物理读时，Oracle会从LRU辅助链表找空闲的BUFFER，然后把LRU辅助的链上的BUFFER挂接到LRU主链的冷端头，实验如下：
首先要保证有LRU辅助链上的BUFFER,即有LRU_FLAG=6或LRU_FLAG=4,如果数据库刚刚启来，可能没有LRU_FLAG=6、LRU_FLAG=6，那需要做大量的物理读操作，才会有LRU_FLAG=6或LRU_FLAG=4
sys@ZMDB> alter system flush buffer_cache;
System altered.
sys@ZMDB> select lru_flag,count(*) from x$bh group by lru_flag;
LRU_FLAG COUNT(*)
---------- ----------
6 208
4 30009
0 2
第一次DUMP整个BUFFER CACHE:
sys@ZMDB> alter session set events‘immediate trace name buffers level 1‘; /u01/app/oracle/diag/rdbms/zmdb/zmdb/trace/zmdb_ora_13480.trc
发生物理读
gyj@ZMDB> conn gyj/gyj
Connected.
gyj@ZMDB> set autot on;
gyj@ZMDB> select id,name, dbms_rowid.rowid_relative_fno(rowid) file#,dbms_rowid.rowid_block_number(rowid) block# from gyj_t1 where id=1;
ID NAME FILE# BLOCK#
---------- ------------------------------ ---------- ----------
1 gyj1 7 139
Execution Plan
----------------------------------------------------------
Plan hash value: 59758809
----------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 14 | 68 (0)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| GYJ_T1 | 1 | 14 | 68 (0)| 00:00:01 |
----------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("ID"=1)
Statistics
----------------------------------------------------------
1 recursive calls
1 db block gets
254 consistent gets
248 physical reads
0 redo size
733 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
sys@ZMDB> select LRU_FLAG,lower(BA),TCH from x$bh where file#=7 and dbablk=139;
LRU_FLAG LOWER(BA) TCH
---------- ---------------- ----------
0 000000007d1b2000 1
4 0000000078558000 0
4 0000000085f68000 0
物理读完成后，再次dump整个buffer cache,
sys@ZMDB> alter session set events‘immediate trace name buffers level 1‘; /u01/app/oracle/diag/rdbms/zmdb/zmdb/trace/zmdb_ora_13511.trc
拿BA=7d1b2000,搜索第一次DUMP的trace文件
/u01/app/oracle/diag/rdbms/zmdb/zmdb/trace/zmdb_ora_13480.trc BH (0x7d3e8098) file#: 3 rdba: 0x00c0586b (3/22635) class: 34 ba: 0x7d1b2000 set: 5 pool: 3 bsz: 8192 bsi: 0 sflg: 1 pwc: 0,25 dbwrid: 0 obj: -1 objn: 0 tsn: 2 afn: 3 hint: f hash: [0x9efa7570,0x9efa7570] lru: [0x7f7f5d30,0x7d3e8050] lru-flags: on_auxiliary_list ckptq: [NULL] fileq: [NULL] objq: [NULL] objaq: [NULL] st: FREE md: NULL fpin: ‘ktuwh03: ktugnb‘ tch: 0 lfb: 33 flags:
拿BA=7d1b2000,搜索第二次DUMP的trace文件
/u01/app/oracle/diag/rdbms/zmdb/zmdb/trace/zmdb_ora_13511.trc BH (0x7d3e8098) file#: 7 rdba: 0x01c0008b (7/139) class: 1 ba: 0x7d1b2000 set: 5 pool: 3 bsz: 8192 bsi: 0 sflg: 1 pwc: 0,25 dbwrid: 0 obj: 22919 objn: 19567 tsn: 7 afn: 7 hint: f hash: [0x787e4bd8,0x9e4cda50] lru: [0x7f7f5d30,0x7d3e8050] ckptq: [NULL] fileq: [NULL] objq: [0x9a88e518,0x7d3e8078] objaq: [0x9a88e508,0x7d3e8088] st: XCURRENT md: NULL fpin: ‘kdswh11: kdst_fetch‘ tch: 1 flags: only_sequential_access LRBA: [0x0.0.0] LSCN: [0x0.0] HSCN: [0xffff.ffffffff] HSUB: [65535]
从上面的两个trace可以得出结论ba: 0x7d1b2000
从lru-flags: on_auxiliary_list(LRU_FLAG=4)到LRU-主链冷端的头部，这个比较特殊在DUMP没有显示LRU_FLAG(LRU_FLAG=0)
观察LRU TCH>=2时冷端移到热端
1、BUFFER手动设为100M
ALTER SYSTEM SET memory_max_target=0 scope=spfile;
ALTER SYSTEM SET memory_target=0;
alter system set sga_target=0;
create table gyj1_t80 (id int,name char(2000));
create table gyj2_t80 (id int,name char(2000));
begin
for i in 1 .. 30000
loop
insert into gyj1_t80 values(i,‘gyj‘||i);
commit;
end loop;
end;
/
SQL> SQL> select bytes/1024/1024||‘M‘ from dba_segments where segment_name=‘GYJ1_T80‘ and owner=‘GYJ‘;
BYTES/1024/1024||‘M‘
-----------------------------------------
80M
begin
for i in 1 .. 30000
loop
insert into gyj2_t80 values(i,‘gyj‘||i);
commit;
end loop;
end;
/
create index idx_gyj1_t80m on gyj1_t80(id);
create index idx_gyj2_t80m on gyj2_t80(id);
SQL> show user;
USER is "GYJ"
SQL> conn / as sysdba
Connected.
SQL> shutdown immediate;
Database closed.
Database dismounted.
ORACLE instance shut down.
SQL> startup
ORACLE instance started.
第一次dump
SQL> alter session set events‘immediate trace name buffers level 1‘;
Session altered.
SQL> select * from v$diag_info where name=‘Default Trace File‘;
INST_ID NAME
---------- --------------------
VALUE
--------------------------------------------------------------------------------
1 Default Trace File
/u01/app/oracle/diag/rdbms/jfdb/jfdb/trace/jfdb_ora_7210.trc
发生一个物理读走索引
set autot on
select id,name,dbms_rowid.rowid_relative_fno(rowid) file#,dbms_rowid.rowid_block_number(rowid) block# from gyj1_t80 where id=1;
SQL> select id,name,dbms_rowid.rowid_relative_fno(rowid) file#,dbms_rowid.rowid_block_number(rowid) block# from gyj1_t80 where id=1;
ID NAME FILE# BLOCK#
---------- -------------------- ---------- ----------
1 gyj1 5 581
select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
SQL> select LRU_FLAG,lower(BA),TCH, decode(state,0,‘free‘,1,‘xcur‘,2,‘scur‘
2 ,3,‘cr‘, 4,‘read‘,5,‘mrec‘,6,‘irec‘,7,‘write‘,8,‘pi‘, 9,‘memory‘,10,‘mwrite‘,11,
3 ‘donated‘, 12,‘protected‘, 13,‘securefile‘, 14,‘siop‘,15,‘recckpt‘, 16, ‘flashf
4 ree‘, 17, ‘flashcur‘, 18, ‘flashna‘) from x$bh where file#=5 and dbablk=581;
LRU_FLAG LOWER(BA) TCH DECODE(STA
---------- ---------------- ---------- ----------
0 000000009fca8000 1 xcur
SQL> select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
LRU_FLAG LOWER(BA) TCH
---------- ---------------- ----------
0 000000009fca8000 5
SQL> set autot traceonly;
SQL> select /*+ index(G) */ count(name) from gyj1_t80 G where id<=8000;
SQL> select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
LRU_FLAG LOWER(BA) TCH
---------- ---------------- ----------
0 000000009fca8000 6
再次发生物理读，此时LRU_FLAG=0变为8，同时TCH=8重置为0
SQL> select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
LRU_FLAG LOWER(BA) TCH
---------- ---------------- ----------
0 000000009fca8000 8
SQL> select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
LRU_FLAG LOWER(BA) TCH
---------- ---------------- ----------
8 000000009fca8000 0
BH (0x9ffe02a8) file#: 5 rdba: 0x01400245 (5/581) class: 1 ba: 0x9fca8000
set: 5 pool: 3 bsz: 8192 bsi: 0 sflg: 2 pwc: 15,19
dbwrid: 0 obj: 13537 objn: 13537 tsn: 5 afn: 5 hint: f
hash: [0xb6a86de0,0xb6a86de0] lru: [0x9ffe0260,0x9ffe9a60]
lru-flags: hot_buffer
ckptq: [NULL] fileq: [NULL] objq: [0x9ffe0618,0x9ffe0028] objaq: [0x9ffe0628,0x9ffe0038]
st: XCURRENT md: NULL fpin: ‘kdswh05: kdsgrp‘ tch: 0
flags:
LRBA: [0x0.0.0] LSCN: [0x0.0] HSCN: [0xffff.ffffffff] HSUB: [65535]
当TCH=0时，再发生大量物理读，地址为9fca8000的BUFFER就被重用了，彻底从BUFFER消失
SQL> select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
LRU_FLAG LOWER(BA) TCH
---------- ---------------- ----------
8 000000009fca8000 0
SQL> select LRU_FLAG,lower(BA),TCH from x$bh where file#=5 and dbablk=581;
no rows selected 通过实验，我们更清楚地了解到物理读LRU的基本流程，可以进一步理解物理读内部的LRU算法。
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