commit cleanouts 0
commit cleanouts successfully completed 0
redo KB read for transport 0
file io wait time 40974
gc cr blocks served 0
transaction tables consistent reads - undo records applied 310
transaction tables consistent read rollbacks 1
cleanouts only - consistent read gets 0
cleanouts and rollbacks - consistent read gets 1
immediate (CR) block cleanout applications 1
deferred (CURRENT) block cleanout applications 0
针对上述测试结果的分析：
Undo header 事务表中的每个slot被覆盖多达10次，session 2还能正常返回scott.tabnow1表修改前的记录，就是利用了undo事务表的回滚机制，为了实现这个回滚：当新的事务开始使用某个undo事务表槽之前会将事务表TRN CTL部分的scn与uba两个值作为ctl max scn、uba字段保存到新事务所拥有的undo block里，也会将这个事务表槽中的scn与dba两个值作为prv tx scn、prev brb字段保存到新事务所拥有的undo block里，例如session 1 中事务使用的uba为8/16 Rec #0x19：
*-----------------------------
* Rec #0x19 slt: 0x1e objn: 90344(0x000160e8) objd: 90344 tblspc: 8(0x00000008)
* Layer: 11 (Row) opc: 1 rci 0x00
Undo type: Regular undo Begin trans Last buffer split: No
Temp Object: No
Tablespace Undo: No
rdba: 0x00000000Ext idx: 0
flg2: 0
*-----------------------------
uba: 0x02000010.187e.18 ctl max scn: 0x0000.00ae9095 prv tx scn: 0x0000.00ae909b
txn start scn: scn: 0x0000.00ae90b2 logon user: 0
prev brb: 33554460 prev bcl: 0
KDO undo record:
KTB Redo
op: 0x04 ver: 0x01
compat bit: 4 (post-11) padding: 1
op: L itl: xid: 0x0002.004.0000111f uba: 0x0200001a.177b.09
flg: C--- lkc: 0 scn: 0x0000.00ad9579
KDO Op code: URP row dependencies Disabled
xtype: XA flags: 0x00000000 bdba: 0x010006f3 hdba: 0x010006f2
itli: 2 ispac: 0 maxfr: 4858
tabn: 0 slot: 0(0x0) flag: 0x2c lock: 0 ckix: 0
ncol: 3 nnew: 1 size: 2
col 0: [ 9] 58 53 24 4e 55 4c 4c 4c 4c