enq:CU的命名规则

霍俊路

你好：
我使用10074dumpOracle 内部锁信息，关于dump文件中的内容不太明白，请指教。
ksqgtl *** CU-b2660a04-00000000 mode=6 flags=0x10 timeout=300 ***
ksqgtl: no transaction
ksqgtl: use existing ksusetxn DID
ksqgtl:
        ksqlkdid: 0001-0013-00000031

*** 2013-01-09 13:02:00.740
*** ksudidTrace: ksqgtl
        ksusesdi:   0000-0000-00000000
        ksusetxn:   0001-0013-00000031
ksqgtl: RETURNS 0

*** 2013-01-09 13:02:00.822
ksqgtl *** CU-b26730e0-00000000 mode=6 flags=0x10 timeout=300 ***
ksqgtl: no transaction
ksqgtl: use existing ksusetxn DID
ksqgtl:
        ksqlkdid: 0001-0013-00000031

*** 2013-01-09 13:02:00.823
*** ksudidTrace: ksqgtl
        ksusesdi:   0000-0000-00000000
        ksusetxn:   0001-0013-00000031
ksqgtl: RETURNS 0

*** 2013-01-09 13:02:00.853
ksqrcl: CU,b26730e0,0
ksqrcl: returns 0

*** 2013-01-09 13:02:00.870
ksqrcl: CU,b2660a04,0
ksqrcl: returns 0


第一个问题： enq:CU的命名规则是什么？它的object id是否对应系统视图中某个字段？
第二个问题：ksqgtl ，ksusesdi，ksusetxn，ksqrcl等4个函数大致的含义能否说明一下。

Maclean Liu(刘相兵

3.4        CU, Bind Enqueue
How Many Resources:
1/child cursor
How Many Locks:
1/child cursor/process.
How Many Users:
processes * open_cursors.
Who Uses:
All processes.
When Used:
This keeps us from hanging on the exclusive pin if somebody else sets the bind types and gets the shared-pin for the execute before we get the exclusive pin for bind (which would be unnecessary at that point). Once we get the enqueue, we’ll check whether the cursor is still unbound, in which case we’ll verify that it was bound the way that we wanted it, or we’ll go ahead and bind it ourselves.
Id1, Id2 Combination:
Child Object Handle, 0.

Maclean Liu(刘相兵

CU

Id1=>Child Object Handle
ID2=>0

ksqgtl get lock
ksqrcl release lock