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052-013题解析

原创 Oracle 作者:pxbibm 时间:2014-03-31 15:44:34 0 删除 编辑
 

13.

You have executed this command to change the size of the database buffer cache:

SQL> ALTER SYSTEM SET DB_CACHE_SIZE=2516582;

System altered.

To verify the change in size, you executed this command:

SQL> SHOW PARAMETER DB_CACHE_SIZE

NAME TYPE VALUE

------------------- ----------- ------------------

db_cache_size big integer 4194304 4M

Why is the value set to 4194304 and not to 2516582?

A.because 4194304 is the granule size

B.because 4194304 is the standard block size

C.because 4194304 is the largest nonstandard block size defined in the database

D.because 4194304 is the total size of data already available in the database buffer cache

Answer: A

答案解析:

本题考得是最小内存管理的颗粒度。
我们可以查询动态系能视图表V$SGAINFO获取最小内存管理的颗粒度。


如果指定的size不是granule size的整数倍,则rounds到整数倍,例如,如果granule size4M,指定DB_CACHE_SIZE as 10 MB,不是4的整数倍,则实际上DB_CACHE_SIZE as 12M


SQL> select * from v$sgainfo;

NAME                                  BYTES RES
-------------------------------- ---------- ---
Fixed SGA Size                      1339740 No
Redo Buffers                        5160960 No
Buffer Cache Size                 213909504 Yes
Shared Pool Size                  314572800 Yes
Large Pool Size                     4194304 Yes
Java Pool Size                      4194304 Yes
Streams Pool Size                         0 Yes
Shared IO Pool Size                       0 Yes
Granule Size                        4194304 No
Maximum SGA Size                  836976640 No
Startup overhead in Shared Pool    58720256 No

NAME                                  BYTES RES
-------------------------------- ---------- ---
Free SGA Memory Available         293601280

12 rows selected.



 

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