这篇文章和之前的《MINSCN与Cache Fusion Read Consistent》 是姊妹篇，他们源于同一个问题帖子。
我们来重现提问者所看到的这样一个场景：
SQL> select * from V$version;
BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
PL/SQL Release 11.2.0.3.0 - Production
CORE 11.2.0.3.0 Production
TNS for Linux: Version 11.2.0.3.0 - Production
NLSRTL Version 11.2.0.3.0 - Production
SQL> select count(*) from gv$instance;
COUNT(*)
----------
2
SQL> select * from global_name;
GLOBAL_NAME
--------------------------------------------------------------------------------
www.askmac.cn
在11gR2 2节点RAC环境中将一个数据块的status修改为XG，假设这个Xcurrent block当前在INSTANCE 2被hold住，这时我们在INSTANCE 1反复查询这个数据块，并观察结果：
SQL> select * from test;
ID
----------
1
2
SQL> select dbms_rowid.rowid_block_number(rowid),dbms_rowid.rowid_relative_fno(rowid) from test;
DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID) DBMS_ROWID.ROWID_RELATIVE_FNO(ROWID)
------------------------------------ ------------------------------------
89233 1
89233 1
SQL> alter system flush buffer_cache;
System altered.
INSTANCE 1 Session A:
SQL> update test set id=id+1 where id=1;
1 row updated.
INSTANCE 1 Session B:
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
1 0
3 1755287
SQL> oradebug setmypid;
Statement processed.
SQL> oradebug dump gc_elements 255;
Statement processed.
SQL> oradebug tracefile_name;
/s01/orabase/diag/rdbms/vprod/VPROD1/trace/VPROD1_ora_19111.trc
GLOBAL CACHE ELEMENT DUMP (address: 0xa4ff3080):
id1: 0x15c91 id2: 0x1 pkey: OBJ#76896 block: (1/89233)
lock: X rls: 0x0 acq: 0x0 latch: 3
flags: 0x20 fair: 0 recovery: 0 fpin: 'kdswh11: kdst_fetch'
bscn: 0x0.146e20 bctx: (nil) write: 0 scan: 0x0
lcp: (nil) lnk: [NULL] lch: [0xa9f6a6f8,0xa9f6a6f8]
seq: 32 hist: 58 145:0 118 66 144:0 192 352 197 48 121 113 424 180 58
LIST OF BUFFERS LINKED TO THIS GLOBAL CACHE ELEMENT:
flg: 0x02000001 lflg: 0x1 state: XCURRENT tsn: 0 tsh: 2
addr: 0xa9f6a5c8 obj: 76896 cls: DATA bscn: 0x0.1ac898
BH (0xa9f6a5c8) file#: 1 rdba: 0x00415c91 (1/89233) class: 1 ba: 0xa9e56000
set: 5 pool: 3 bsz: 8192 bsi: 0 sflg: 3 pwc: 0,15
dbwrid: 0 obj: 76896 objn: 76896 tsn: 0 afn: 1 hint: f
hash: [0x91f4e970,0xbae9d5b8] lru: [0x91f58848,0xa9f6a828]
lru-flags: debug_dump
obj-flags: object_ckpt_list
ckptq: [0x9df6d1d8,0xa9f6a740] fileq: [0xa2ece670,0xbdf4ed68] objq: [0xb4964e00,0xb4964e00] objaq: [0xb4964de0,0xb4964de0]
st: XCURRENT md: NULL fpin: 'kdswh11: kdst_fetch' tch: 2 le: 0xa4ff3080
flags: buffer_dirty redo_since_read
LRBA: [0x19.5671.0] LSCN: [0x0.1ac898] HSCN: [0x0.1ac898] HSUB: [1]
buffer tsn: 0 rdba: 0x00415c91 (1/89233)
scn: 0x0000.001ac898 seq: 0x01 flg: 0x00 tail: 0xc8980601
frmt: 0x02 chkval: 0x0000 type: 0x06=trans data
可以看到此时block: (1/89233)的GLOBAL CACHE ELEMENT DUMP中LOCK状态仍是X 而非XG ， 这是因为这个Current Block仅在一个Instance中被modify修改过，没有在全局范围内被更新过。
紧接着在Instance 2 修改该块：
Instance 2 Session C:
SQL> update test set id=id+1 where id=2;
1 row updated.
Instance 2 Session D:
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
1 0
3 1756658
SQL> oradebug setmypid;
Statement processed.
SQL> oradebug dump gc_elements 255;
Statement processed.
SQL> oradebug tracefile_name;
/s01/orabase/diag/rdbms/vprod/VPROD2/trace/VPROD2_ora_13038.trc
GLOBAL CACHE ELEMENT DUMP (address: 0x89fb25a0):
id1: 0x15c91 id2: 0x1 pkey: OBJ#76896 block: (1/89233)
lock: XG rls: 0x0 acq: 0x0 latch: 3
flags: 0x20 fair: 0 recovery: 0 fpin: 'kduwh01: kdusru'
bscn: 0x0.1acdf3 bctx: (nil) write: 0 scan: 0x0
lcp: (nil) lnk: [NULL] lch: [0x96f4cf80,0x96f4cf80]
seq: 61 hist: 324 21 143:0 19 16 352 329 144:6 14 7 352 197
LIST OF BUFFERS LINKED TO THIS GLOBAL CACHE ELEMENT:
flg: 0x0a000001 state: XCURRENT tsn: 0 tsh: 1
addr: 0x96f4ce50 obj: 76896 cls: DATA bscn: 0x0.1acdf6
BH (0x96f4ce50) file#: 1 rdba: 0x00415c91 (1/89233) class: 1 ba: 0x96bd4000
set: 5 pool: 3 bsz: 8192 bsi: 0 sflg: 2 pwc: 0,15
dbwrid: 0 obj: 76896 objn: 76896 tsn: 0 afn: 1 hint: f
hash: [0x96ee1fe8,0xbae9d5b8] lru: [0x96f4d0b0,0x96f4cdc0]
obj-flags: object_ckpt_list
ckptq: [0xbdf519b8,0x96f4d5a8] fileq: [0xbdf519d8,0xbdf519d8] objq: [0xb4a47b90,0xb4a47b90] objaq: [0x96f4d0e8,0xb4a47b70]
st: XCURRENT md: NULL fpin: 'kduwh01: kdusru' tch: 1 le: 0x89fb25a0
flags: buffer_dirty redo_since_read remote_transfered
LRBA: [0x11.9e18.0] LSCN: [0x0.1acdf6] HSCN: [0x0.1acdf6] HSUB: [1]
buffer tsn: 0 rdba: 0x00415c91 (1/89233)
scn: 0x0000.001acdf6 seq: 0x01 flg: 0x00 tail: 0xcdf60601
frmt: 0x02 chkval: 0x0000 type: 0x06=trans data
GCS CLIENT 0x89fb2618,6 resp[(nil),0x15c91.1] pkey 76896.0
grant 2 cvt 0 mdrole 0x42 st 0x100 lst 0x20 GRANTQ rl G0
master 1 owner 2 sid 0 remote[(nil),0] hist 0x94121c601163423c
history 0x3c.0x4.0xd.0xb.0x1.0xc.0x7.0x9.0x14.0x1.
cflag 0x0 sender 1 flags 0x0 replay# 0 abast (nil).x0.1 dbmap (nil)
disk: 0x0000.00000000 write request: 0x0000.00000000
pi scn: 0x0000.00000000 sq[(nil),(nil)]
msgseq 0x1 updseq 0x0 reqids[6,0,0] infop (nil) lockseq x2b8
pkey 76896.0
hv 93 [stat 0x0, 1->1, wm 32768, RMno 0, reminc 18, dom 0]
kjga st 0x4, step 0.0.0, cinc 20, rmno 6, flags 0x0
lb 0, hb 0, myb 15250, drmb 15250, apifrz 0
在Instance 2中被修改过后block: (1/89233)的 GLOBAL CACHE ELEMENT Lock Convert成lock: XG
除了通过GC_ELEMENTS DUMP来分析XCUR Cache Fusion外，也可以直接查询X$ VIEW，主要是 X$LE X$KJBR X$KJBL， 这三个X$ VIEW的更多信息可以很方便地从我的博客中找到：
INSTANCE 2 Session D:
SELECT *
FROM x$le
WHERE le_addr IN (SELECT le_addr
FROM x$bh
WHERE obj IN (SELECT data_object_id
FROM dba_objects
WHERE owner = 'SYS'
AND object_name = 'TEST')
AND class = 1
AND state != 3);
ADDR INDX INST_ID LE_ADDR LE_ID1 LE_ID2
---------------- ---------- ---------- ---------------- ---------- ----------
LE_RLS LE_ACQ LE_FLAGS LE_MODE LE_WRITE LE_LOCAL LE_RECOVERY
---------- ---------- ---------- ---------- ---------- ---------- -----------
LE_BLKS LE_TIME LE_KJBL
---------- ---------- ----------------
00007F94CA14CF60 7003 2 0000000089FB25A0 89233 1
0 0 32 2 0 1 0
1 0 0000000089FB2618
PCM Resource NAME由[ID1][ID2],[BL]等组成， ID1和ID2 通过blockno和 fileno计算获得， 这里我们直接参考以上GC_elements dump中的 id1: 0x15c91 id2: 0x1 pkey: OBJ#76896 block: (1/89233)信息 ，则 kjblname 和 kjbrname 应以”[0x15c91][0x1],[BL]” 开头：
INSTANCE 2 Session D:
SQL> set linesize 80 pagesize 1400
SQL> SELECT *
2 FROM x$kjbl l
3 WHERE l.kjblname LIKE '%[0x15c91][0x1],[BL]%';
ADDR INDX INST_ID KJBLLOCKP KJBLGRANT KJBLREQUE
---------------- ---------- ---------- ---------------- --------- ---------
KJBLROLE KJBLRESP KJBLNAME
---------- ---------------- ------------------------------
KJBLNAME2 KJBLQUEUE
------------------------------ ----------
KJBLLOCKST KJBLWRITING
---------------------------------------------------------------- -----------
KJBLREQWRITE KJBLOWNER KJBLMASTER KJBLBLOCKED KJBLBLOCKER KJBLSID KJBLRDOMID
------------ ---------- ---------- ----------- ----------- ---------- ----------
KJBLPKEY
----------
00007F94CA22A288 451 2 0000000089FB2618 KJUSEREX KJUSERNL
0 00 [0x15c91][0x1],[BL][ext 0x0,0x
89233,1,BL 0
GRANTED 0
0 1 0 0 0 0 0
76896
SQL> SELECT r.* FROM x$kjbr r WHERE r.kjbrname LIKE '%[0x15c91][0x1],[BL]%';
no rows selected
Instance 1 session B:
SQL> SELECT r.* FROM x$kjbr r WHERE r.kjbrname LIKE '%[0x15c91][0x1],[BL]%';
ADDR INDX INST_ID KJBRRESP KJBRGRANT KJBRNCVL
---------------- ---------- ---------- ---------------- --------- ---------
KJBRROLE KJBRNAME KJBRMASTER KJBRGRANTQ
---------- ------------------------------ ---------- ----------------
KJBRCVTQ KJBRWRITER KJBRSID KJBRRDOMID KJBRPKEY
---------------- ---------------- ---------- ---------- ----------
00007F801ACA68F8 1355 1 00000000B5A62AE0 KJUSEREX KJUSERNL
0 [0x15c91][0x1],[BL][ext 0x0,0x 0 00000000B48BB330
00 00 0 0 76896
接着我们将在Instance 1上查询block: (1/89233)，这应当会引发Instance 2 build cr block 并传输给Instance 1， 为了进一步了解其细节 我们分别对 Instance 1的 Foreground Process 和 Instance 2的LMS进程做详细的RAC TRACE:
Instance 2:
[oracle@vrh2 ~]$ ps -ef|grep ora_lms|grep -v grep
oracle 23364 1 0 Apr29 ? 00:33:15 ora_lms0_VPROD2
SQL> oradebug setospid 23364
Oracle pid: 13, Unix process pid: 23364, image: oracle@vrh2.oracle.com (LMS0)
SQL> oradebug event 10046 trace name context forever,level 8:10708 trace name context forever,level 103: trace[rac.*] disk high;
Statement processed.
SQL> oradebug tracefile_name
/s01/orabase/diag/rdbms/vprod/VPROD2/trace/VPROD2_lms0_23364.trc
Instance 1 session B :
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
3 1756658
3 1756661
3 1755287
Instance 1 session A :
SQL> alter session set events '10046 trace name context forever,level 8:10708 trace name context forever,level 103: trace[rac.*] disk high';
Session altered.
SQL> select * from test;
ID
----------
2
2
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
3 1761520
从x$BH的信息来看，如我们所预期的Instance 1收到了build好的CR block，进一步分析 TRACE 文件：
Instance 1 foreground Process: PARSING IN CURSOR #140336527348792 len=18 dep=0 uid=0 oct=3 lid=0 tim=1335939136125254 hv=1689401402 ad='b1a4c828' sqlid='c99yw1xkb4f1u' select * from test END OF STMT PARSE #140336527348792:c=2999,e=2860,p=0,cr=0,cu=0,mis=1,r=0,dep=0,og=1,plh=1357081020,tim=1335939136125253 EXEC #140336527348792:c=0,e=40,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1357081020,tim=1335939136125373 WAIT #140336527348792: nam='SQL*Net message to client' ela= 6 driver id=1650815232 #bytes=1 p3=0 obj#=0 tim=1335939136125420 *** 2012-05-02 02:12:16.125 kclscrs: req=0 block=1/89233 2012-05-02 02:12:16.125574 : kjbcro[0x15c91.1 76896.0][4] *** 2012-05-02 02:12:16.125 kclscrs: req=0 typ=nowait-abort *** 2012-05-02 02:12:16.125 kclscrs: bid=1:3:1:0:f:1e:0:0:10:0:0:0:1:2:4:1:20:0:0:0:c3:49:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:4:3:2:1:2:0:1c:0:4d:26:a3:52:0:0:0:0:c7:c:ca:62:c3:49:0:0:0:0:1:0:14:8e:47:76:1:2:dc:5:a9:fe:17:75:0:0:0:0:0:0:0:0:0:0:0:0:99:ed:0:0:0:0:0:0:10:0:0:0 2012-05-02 02:12:16.125718 : kjbcro[0x15c91.1 76896.0][4] 2012-05-02 02:12:16.125751 : GSIPC:GMBQ: buff 0xba0ee018, queue 0xbb79a7b8, pool 0x60013fa0, freeq 0, nxt 0xbb79a7b8, prv 0xbb79a7b8 2012-05-02 02:12:16.125780 : kjbsentscn[0x0.1ae0f0][to 2] 2012-05-02 02:12:16.125806 : GSIPC:SENDM: send msg 0xba0ee088 dest x20001 seq 177740 type 36 tkts xff0000 mlen x1680198 2012-05-02 02:12:16.125918 : kjbmscr(0x15c91.1)reqid=0x8(req 0xa4ff30f8)(rinst 1)hldr 2(infosz 200)(lseq x2b8) 2012-05-02 02:12:16.126959 : GSIPC:KSXPCB: msg 0xba0ee088 status 30, type 36, dest 2, rcvr 1 *** 2012-05-02 02:12:16.127 kclwcrs: wait=0 tm=1233 *** 2012-05-02 02:12:16.127 kclwcrs: got 1 blocks from ksxprcv WAIT #140336527348792: nam='gc cr block 2-way' ela= 1233 p1=1 p2=89233 p3=1 obj#=76896 tim=1335939136127199 2012-05-02 02:12:16.127272 : kjbcrcomplete[0x15c91.1 76896.0][0] 2012-05-02 02:12:16.127309 : kjbrcvdscn[0x0.1ae0f0][from 2][idx 2012-05-02 02:12:16.127329 : kjbrcvdscn[no bscn <= rscn 0x0.1ae0f0][from 2]
前台进程 kjbcro[0x15c91.1 76896.0][4] kjbsentscn[0x0.1ae0f0][to 2] 向Instance 2请求SCN=1ae0f0=1761520的 block: (1/89233)，并进入’gc cr block 2-way’ 等待，之后成功收到请求的CR block。
Instance 2 LMS TRACE
2012-05-02 02:12:15.634057 : GSIPC:RCVD: ksxp msg 0x7f16e1598588 sndr 1 seq 0.177740 type 36 tkts 0 2012-05-02 02:12:15.634094 : GSIPC:RCVD: watq msg 0x7f16e1598588 sndr 1, seq 177740, type 36, tkts 0 2012-05-02 02:12:15.634108 : GSIPC:TKT: collect msg 0x7f16e1598588 from 1 for rcvr -1, tickets 0 2012-05-02 02:12:15.634162 : kjbrcvdscn[0x0.1ae0f0][from 1][idx 2012-05-02 02:12:15.634186 : kjbrcvdscn[no bscn1, wm 32768, RMno 0, reminc 18, dom 0] kjga st 0x4, step 0.0.0, cinc 20, rmno 6, flags 0x0 lb 0, hb 0, myb 15250, drmb 15250, apifrz 0 GCS CLIENT END 2012-05-02 02:12:15.635211 : kjbdowncvt[0x15c91.1 76896.0][1][options x0] 2012-05-02 02:12:15.635230 : GSIPC:AMBUF: rcv buff 0x7f16e1c56420, pool rcvbuf, rqlen 1103 2012-05-02 02:12:15.635308 : GSIPC:GPBMSG: new bmsg 0x7f16e1c56490 mb 0x7f16e1c56420 msg 0x7f16e1c564b0 mlen 152 dest x101 flushsz -1 2012-05-02 02:12:15.635334 : kjbmslset(0x15c91.1)) seq 0x4 reqid=0x6 (shadow 0xb48bb330.xb)(rsn 2)(mas@1) 2012-05-02 02:12:15.635355 : GSIPC:SPBMSG: send bmsg 0x7f16e1c56490 blen 184 msg 0x7f16e1c564b0 mtype 33 attr|dest x30101 bsz|fsz x1ffff 2012-05-02 02:12:15.635377 : GSIPC:SNDQ: enq msg 0x7f16e1c56490, type 65521 seq 118669, inst 1, receiver 1, queued 1 *** 2012-05-02 02:12:15.635 kclccctx: cleanup copy 0x7f16e1d94798 2012-05-02 02:12:15.635479 : [kjmpmsgi:compl][type 36][msg 0x7f16e1598588][seq 177740.0][qtime 0][ptime 1257] 2012-05-02 02:12:15.635511 : GSIPC:BSEND: flushing sndq 0xb491dd28, id 1, dcx 0xbc516778, inst 1, rcvr 1 qlen 0 1 2012-05-02 02:12:15.635536 : GSIPC:BSEND: no batch1 msg 0x7f16e1c56490 type 65521 len 184 dest (1:1) 2012-05-02 02:12:15.635557 : kjbsentscn[0x0.1ae0f1][to 1] 2012-05-02 02:12:15.635578 : GSIPC:SENDM: send msg 0x7f16e1c56490 dest x10001 seq 118669 type 65521 tkts x10002 mlen xb800e8 WAIT #0: nam='gcs remote message' ela= 180 waittime=1 poll=0 event=0 obj#=0 tim=1335939135635819 2012-05-02 02:12:15.635853 : GSIPC:RCVD: ksxp msg 0x7f16e167e0b0 sndr 1 seq 0.177741 type 32 tkts 0 2012-05-02 02:12:15.635875 : GSIPC:RCVD: watq msg 0x7f16e167e0b0 sndr 1, seq 177741, type 32, tkts 0 2012-05-02 02:12:15.636012 : GSIPC:TKT: collect msg 0x7f16e167e0b0 from 1 for rcvr -1, tickets 0 2012-05-02 02:12:15.636040 : kjbrcvdscn[0x0.1ae0f1][from 1][idx 2012-05-02 02:12:15.636060 : kjbrcvdscn[no bscn <= rscn 0x0.1ae0f1][from 1] 2012-05-02 02:12:15.636082 : GSIPC:TKT: dest (1:1) rtkt not acked 1 unassigned bufs 0 tkts 0 newbufs 0 2012-05-02 02:12:15.636102 : GSIPC:TKT: remove ctx dest (1:1) 2012-05-02 02:12:15.636125 : [kjmxmpm][type 32][seq 0.177741][msg 0x7f16e167e0b0][from 1] 2012-05-02 02:12:15.636146 : kjbmpocr(0xb0.6)seq 0x1,reqid=0x23a,(client 0x9fff7b58,0x1)(from 1)(lseq xdf0)
2号实例上LMS进程的情况是这样的 收到gcs remote message GSIPC 关于要求SCN=[0x0.1ae0f0] block=1/89233的请求，进入BAST kjbmpbast(0x15c91.1)，因为 block=1/89233已被写入到磁盘 触发fairness算法(在11.2.0.3中默认_fairness_threshold=2)，对current block做KCL: F156: fairness downconvert，从Xcurrent DownConvert为 Scurrent:
Instance 2: SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0; STATE CR_SCN_BAS ---------- ---------- 2 0 3 1756658
之后Instance 2 LMS 将cr block加入到 kjbmslset(0x15c91.1)) 传送队列SEND QUEUE GSIPC:SNDQ: enq msg 0x7f16e1c56490。
接着我们再次在Instance 1上运行对 block: (1/89233)的查询 看会发生什么：
Instance 2:
SQL> select CURRENT_RESULTS,LIGHT_WORKS from v$cr_block_server;
CURRENT_RESULTS LIGHT_WORKS
--------------- -----------
29273 437
Instance 1 session A:
SQL>
SQL> select * from test;
ID
----------
2
2
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
3 1761942
3 1761932
1 0
3 1761520
Instance 2:
SQL> select CURRENT_RESULTS,LIGHT_WORKS from v$cr_block_server;
CURRENT_RESULTS LIGHT_WORKS
--------------- -----------
29274 437
select * from test
END OF STMT
PARSE #140336529675592:c=0,e=337,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1357081020,tim=1335939668940051
EXEC #140336529675592:c=0,e=96,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1357081020,tim=1335939668940204
WAIT #140336529675592: nam='SQL*Net message to client' ela= 5 driver id=1650815232 #bytes=1 p3=0 obj#=0 tim=1335939668940348
*** 2012-05-02 02:21:08.940
kclscrs: req=0 block=1/89233
2012-05-02 02:21:08.940676 : kjbcro[0x15c91.1 76896.0][5]
*** 2012-05-02 02:21:08.940
kclscrs: req=0 typ=nowait-abort
*** 2012-05-02 02:21:08.940
kclscrs: bid=1:3:1:0:f:21:0:0:10:0:0:0:1:2:4:1:20:0:0:0:c3:49:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:0:4:3:2:1:2:0:1f:0:4d:26:a3:52:0:0:0:0:c7:c:ca:62:c3:49:0:0:0:0:1:0:17:8e:47:76:1:2:dc:5:a9:fe:17:75:0:0:0:0:0:0:0:0:0:0:0:0:99:ed:0:0:0:0:0:0:10:0:0:0
2012-05-02 02:21:08.940799 : kjbcro[0x15c91.1 76896.0][5]
2012-05-02 02:21:08.940833 : GSIPC:GMBQ: buff 0xba0ee018, queue 0xbb79a7b8, pool 0x60013fa0, freeq 0, nxt 0xbb79a7b8, prv 0xbb79a7b8
2012-05-02 02:21:08.940859 : kjbsentscn[0x0.1ae28c][to 2]
2012-05-02 02:21:08.940870 : GSIPC:SENDM: send msg 0xba0ee088 dest x20001 seq 177810 type 36 tkts xff0000 mlen x1680198
2012-05-02 02:21:08.940976 : kjbmscr(0x15c91.1)reqid=0xa(req 0xa4ff30f8)(rinst 1)hldr 2(infosz 200)(lseq x2b8)
2012-05-02 02:21:08.941314 : GSIPC:KSXPCB: msg 0xba0ee088 status 30, type 36, dest 2, rcvr 1
*** 2012-05-02 02:21:08.941
kclwcrs: wait=0 tm=707
*** 2012-05-02 02:21:08.941
kclwcrs: got 1 blocks from ksxprcv
2012-05-02 02:21:08.941818 : kjbassume[0x15c91.1][sender 2][mymode x1][myrole x0][srole x0][flgs x0][spiscn 0x0.0][swscn 0x0.0]
2012-05-02 02:21:08.941852 : kjbrcvdscn[0x0.1ae28d][from 2][idx 2012-05-02 02:21:08.941871 : kjbrcvdscn[no bscn
前台进程与上一次类似申请的是SCN=[0x0.1ae28c]=1761932 Version的CR block， 但这一次receive的居然是Xcurrent Block且其SCN=1ae28d=1761933，Instance 1收到了Xcurrent Block后自己build出了查询所需要的SCN=1761932的CR BLOCK， 因为实际收到的是Current block，所以等待事件变成了’gc current block 2-way’。
这里可以看到前台进程并没有request current block，而仍是使用kjbcro；追根述源是Instance 2的LMS进程给它传送了Current Block：
Instance 2 LMS trace: 2012-05-02 02:21:08.448743 : GSIPC:RCVD: ksxp msg 0x7f16e14a4398 sndr 1 seq 0.177810 type 36 tkts 0 2012-05-02 02:21:08.448778 : GSIPC:RCVD: watq msg 0x7f16e14a4398 sndr 1, seq 177810, type 36, tkts 0 2012-05-02 02:21:08.448798 : GSIPC:TKT: collect msg 0x7f16e14a4398 from 1 for rcvr -1, tickets 0 2012-05-02 02:21:08.448816 : kjbrcvdscn[0x0.1ae28c][from 1][idx 2012-05-02 02:21:08.448834 : kjbrcvdscn[no bscn <= rscn 0x0.1ae28c][from 1] 2012-05-02 02:21:08.448857 : GSIPC:TKT: dest (1:1) rtkt not acked 2 unassigned bufs 0 tkts 0 newbufs 0 2012-05-02 02:21:08.448875 : GSIPC:TKT: remove ctx dest (1:1) 2012-05-02 02:21:08.448970 : [kjmxmpm][type 36][seq 0.177810][msg 0x7f16e14a4398][from 1] 2012-05-02 02:21:08.448993 : kjbmpbast(0x15c91.1) reqid=0x6 (req 0xa4ff30f8)(reqinst 1)(reqid 10)(flags x0) *** 2012-05-02 02:21:08.449 kclcrrf: req=48054 block=1/89233 *** 2012-05-02 02:21:08.449 kcl_compress_block: compressed: 6 free space: 7680 2012-05-02 02:21:08.449085 : kjbsentscn[0x0.1ae28d][to 1] 2012-05-02 02:21:08.449142 : kjbdeliver[to 1][0xa4ff30f8][10][current 1] 2012-05-02 02:21:08.449164 : kjbmssch(reqlock 0xa4ff30f8,10)(to 1)(bsz 344) 2012-05-02 02:21:08.449183 : GSIPC:AMBUF: rcv buff 0x7f16e18bcec8, pool rcvbuf, rqlen 1102 *** 2012-05-02 02:21:08.449 kclccctx: cleanup copy 0x7f16e1d94838 *** 2012-05-02 02:21:08.449 kcltouched: touch seconds 3271 *** 2012-05-02 02:21:08.449 kclgrantlk: req=48054 2012-05-02 02:21:08.449347 : [kjmpmsgi:compl][type 36][msg 0x7f16e14a4398][seq 177810.0][qtime 0][ptime 1119] WAIT #0: nam='gcs remote message' ela= 568 waittime=1 poll=0 event=0 obj#=0 tim=1335939668449962 2012-05-02 02:21:08.450001 : GSIPC:RCVD: ksxp msg 0x7f16e1bb22a0 sndr 1 seq 0.177811 type 32 tkts 0 2012-05-02 02:21:08.450024 : GSIPC:RCVD: watq msg 0x7f16e1bb22a0 sndr 1, seq 177811, type 32, tkts 0 2012-05-02 02:21:08.450043 : GSIPC:TKT: collect msg 0x7f16e1bb22a0 from 1 for rcvr -1, tickets 0 2012-05-02 02:21:08.450060 : kjbrcvdscn[0x0.1ae28e][from 1][idx 2012-05-02 02:21:08.450078 : kjbrcvdscn[no bscn <= rscn 0x0.1ae28e][from 1] 2012-05-02 02:21:08.450097 : GSIPC:TKT: dest (1:1) rtkt not acked 3 unassigned bufs 0 tkts 0 newbufs 0 2012-05-02 02:21:08.450116 : GSIPC:TKT: remove ctx dest (1:1) 2012-05-02 02:21:08.450136 : [kjmxmpm][type 32][seq 0.177811][msg 0x7f16e1bb22a0][from 1] 2012-05-02 02:21:08.450155 : kjbmpocr(0xb0.6)seq 0x1,reqid=0x23e,(client 0x9fff7b58,0x1)(from 1)(lseq xdf4)
为什么Instance 2上的LMS要偷懒，不构造build cr block，而直接传输给Instance 1自己所有的Current Block呢？通过观察Instance 2上的v$cr_block_server视图可以发现LIGHT_WORKS字段在发生current block transfer后并没有增长，说明并没有触发 CR server的 Light Work Rule(Light Work Rule是8i Cr Server中就存在的优化算法，当Remote LMS创建 build CR涉及的工作过多时，resource holder的LMS会选择传输现有block，而不完成CR build If creating the consistent read version block involves too much work (such as reading blocks from disk), then the holder sends the block to the requestor, and the requestor completes the CR fabrication. The holder maintains a fairness counter of CR requests. After the fairness threshold is reached, the holder downgrades it to lock mode.)。
到底何种条件下 CR Request 才会收到Current Block呢？
答案是：针对不同种类class的block，CR server有不同的做法。 对于undo block或者 undo header block的CR quest， LMS总是传输Current Block， 这是为了给 远程节点 可能正在进行的 block cleanout、 CR Version构造提供便利。
对于普通的数据块 data blocks， 默认情况下总是 CR quest & CR received的(排除之前所说的Light Work Rule，LMS”偷懒”)， 除非Current Block已经DownConvert降级为S lock，那么LMS进程会宣称直接ship一个current version的block。
为了证明这一点 ， 我们再次测试 ，这次我们将控制DownConvert的隐藏参数”_fairness_threshold“提高到200，这将造成Xcurrent Block无法降级为Scurrent， 也将导致LMS更少地传输Current Version的Data Block:
SQL> show parameter fair
NAME TYPE VALUE
------------------------------------ ----------- ------------------------------
_fairness_threshold integer 200
Instance 1:
SQL> update test set id=id+1 where id=4;
1 row updated.
Instance 2:
SQL> update test set id=id+1 where id=2;
1 row updated.
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
1 0
3 1838166
在Instance 1上 不断查询，并 通过instance 2的 v$cr_block_server验证
instance 1
SQL> select * from test;
ID
----------
10
3
instance 2:
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
1 0
3 1883707
8 0
SQL> select * from test;
ID
----------
10
3
SQL> select state,cr_scn_bas from x$bh where file#=1 and dbablk=89233 and state!=0;
STATE CR_SCN_BAS
---------- ----------
1 0
3 1883707
8 0
...................
SQL> /
STATE CR_SCN_BAS
---------- ----------
2 0
3 1883707
3 1883695
repeat cr request on Instance 1
SQL> /
STATE CR_SCN_BAS
---------- ----------
8 0
3 1883707
3 1883695
实际测试发现_fairness_threshold似乎存在某个上限，虽然设置为200 但是在不多的几次CR serve后就Downgrade了lock， 这造成对data block的 CR Request最终还是Receive了 Current Block。
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