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The governance, the protocols, and the topology of the Internet during the period from 1983 through 1995 underwent equally drastic change The National Science Foundation (NSF) along with the NSFnet dominated the Internet s topological landscape from 1986 to 1995, not unlike the Roman Senate exerting its influence during the republican period in Rome However, the balance of power on the Internet was well maintained by the emergence and interconnection of the NSFnet with numerous regional networks (MICHnet, SURAnet, BARRnet, PREPnet, MOREnet, and more) Those were the Roman plebeians keeping the patrician-dominated Senate in check! While no single business entity or institution emerged in 1995 to govern the Internet on the order of the Roman emperors, the geographical breadth and scope of the empire set the pattern for the interconnection of Autonomous Systems (ASs) comprising the modern-day Internet.

would have about 17 background processes started up ops$tkyte%ORA11GR2> select paddr, name, description 2 from v$bgprocess 3 where paddr <> '00' 4 order by paddr desc 5 / PADDR -------32AF0E64 32AEF8B4 32AEE304 32AED82C 32AECD54 32AEC27C 32AEB7A4 32AEACCC 32AEA1F4 32AE971C 32AE8C44 32AE816C 32AE7694 32AE6BBC 32AE60E4 32AE560C 32AE4B34 NAME ---CJQ0 QMNC MMNL MMON RECO SMON CKPT LGWR DBW0 MMAN DIA0 PSP0 DBRM DIAG GEN0 VKTM PMON DESCRIPTION ---------------------------------------------------------------Job Queue Coordinator AQ Coordinator Manageability Monitor Process 2 Manageability Monitor Process distributed recovery System Monitor Process checkpoint Redo etc. db writer process 0 Memory Manager diagnosibility process 0 process spawner 0 DataBase Resource Manager diagnosibility process generic0 Virtual Keeper of TiMe process process cleanup

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17 rows selected. Using the same init.ora, only replacing memory_target with sga_target and pga_aggregate_target, in Oracle Database 10g Release 2, you might only see 12: ops$tkyte %ORA10GR2> select paddr, name, description 2 from v$bgprocess 3 where paddr <> '00' 4 order by paddr desc 5 / PADDR -------23D27AC4 23D27508 23D26990 23D263D4 23D25E18 23D2585C 23D252A0 23D24CE4 23D24728 23D2416C 23D23BB0 23D235F4 NAME ---CJQ0 QMNC MMNL MMON RECO SMON CKPT LGWR DBW0 MMAN PSP0 PMON DESCRIPTION ---------------------------------------------------------------Job Queue Coordinator AQ Coordinator Manageability Monitor Process 2 Manageability Monitor Process distributed recovery System Monitor Process checkpoint Redo etc. db writer process 0 Memory Manager process spawner 0 process cleanup

Compile() and DynamicInvoke() are expensive operations to conduct. You don t want to be doing these operations on a per-item basis as you search through a data structure.

VARCHAR2(20) instead of a NUMBER(4) type, we may use the query select cast( empno as VARCHAR2(20) ) e from emp We re now ready to query the data Let s see what one row might look like: ops$tkyte@ORA11GR2> select deptno, dname, loc, demps AS employees 2 from dept_and_emp d 3 where deptno = 10 4 / DEPTNO DNAME LOC EMPLOYEES(EMPNO, ENAME, JOB, ---------- -------------- ------------- ---------------------------10 ACCOUNTING NEW YORK EMP_TAB_TYPE(EMP_TYPE(7782, 'CLARK', 'MANAGER', 7839, '0 9-JUN-81', 2450, NULL), EMP_ TYPE(7839, 'KING', 'PRESIDEN T', NULL, '17-NOV-81', 5000, NULL), EMP_TYPE(7934, 'MILL ER', 'CLERK', 7782, '23-JAN82', 1300, NULL)) All of the data is there in a single column Most applications, unless they are specifically written for the object-relational features, will not be able to deal with this particular column.

For example, ODBC doesn t have a way to deal with a nested table (JDBC, OCI, Pro*C, PL/SQL, and most other APIs and languages do) For those cases, Oracle provides a way to un-nest a collection and treat it much like a relational table: ops$tkyte@ORA11G2> select ddeptno, ddname, emp* 2 from dept_and_emp D, table(d.

Determining whether you have an index is much simpler: Private Function HasIndexFor(ByVal [property] As String) As Boolean _ Implements IIndexSet(Of T)HasIndexFor Return _internalIndexSetContainsKey([property]) End Function Private Function HasIndexablePropertyOnLeft( _ ByVal leftSide As Expression) As Boolean If leftSideNodeType = ExpressionTypeMemberAccess Then Return TryCast(Me, IIndexSet(Of T))HasIndexFor(DirectCast(leftSide, _ MemberExpression)MemberName) Else Return False End If End Function Determining whether you have an index for a given property specified by a string is relatively simple, because you have an internal field _internalIndexSet that is really just a Dictionary of indices You can call ContainsKey() on the _internalIndexSet to determine whether you have the index The examination of the expression that represents the left side of the total expression is slightly more complex You need for this left-side expression to be a MemberExpression in order to make any sense of it.

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