关于DBA的换算问题，我们知道了
10bit的file number，每个对象最多有1022个文件（2个文件预留）
22bit的block number，每个文件最多有4M个BLOCK
及DBA中高10位是文件号，第22位是块号，一般我们使用dbms_utility进行转换，但是
有了算法就可以模拟一个换算的C语言小程序，利用C语言位域结构体
其实我们只要加权相加即可：
#include
#include
#include
#include
#define dn data->d
struct db_addr
{
unsigned d1:1;
unsigned d2:1;
unsigned d3:1;
unsigned d4:1;
unsigned d5:1;
unsigned d6:1;
unsigned d7:1;
unsigned d8:1;
unsigned d9:1;
unsigned d10:1;
unsigned d11:1;
unsigned d12:1;
unsigned d13:1;
unsigned d14:1;
unsigned d15:1;
unsigned d16:1;
unsigned d17:1;
unsigned d18:1;
unsigned d19:1;
unsigned d20:1;
unsigned d21:1;
unsigned d22:1;
unsigned d23:1;
unsigned d24:1;
unsigned d25:1;
unsigned d26:1;
unsigned d27:1;
unsigned d28:1;
unsigned d29:1;
unsigned d30:1;
unsigned d31:1;
unsigned d32:1;
};
void main ()
{
long dba,block1,block2,block3,block4,block5,file1,file2;
long blocksum,filesum;
struct db_addr *data;
printf("please input X dba!\n");
scanf("%x",&dba);
data=&dba;
block1=data->d1+data->d2*2+data->d3*pow(2,2)+data->d4*pow(2,3)+data->d5*pow(2,4);
// printf("%d",data->d1);
block2=data->d6*pow(2,5)+data->d7*pow(2,6)+data->d8*pow(2,7)+data->d9*pow(2,8)+data->d10*pow(2,9);
block3=data->d11*pow(2,10)+data->d12*pow(2,11)+data->d13*pow(2,12)+data->d14*pow(2,13)+data->d15*pow(2,14);
block4=data->d16*pow(2,15)+data->d17*pow(2,17)+data->d18*pow(2,17)+data->d19*pow(2,18)+data->d20*pow(2,19);
block5=data->d21*pow(2,20)+data->d22*pow(2,21);
blocksum=block1+block2+block3+block4+block5;
file1=data->d23*pow(2,0)+data->d24*pow(2,1)+data->d25*pow(2,2)+data->d26*pow(2,3)+data->d27*pow(2,4);
file2=data->d28*pow(2,5)+data->d29*pow(2,6)+data->d30*pow(2,7)+data->d31*pow(2,8)+data->d32*pow(2,9);
filesum=file1+file2;
printf("file id is:%ld\nblocks id is :%ld\n",filesum,blocksum);
}
如果我们使用程序换算
2c000d3
如下：
please input X dba!
2c000d3
file id is:11
blocks id is :211
实际和dbms_utility得到的是相同的。
SQL> select to_number('2c000d3','xxxxxxxxxxxxxx') from dual;
TO_NUMBER('2C000D3','XXXXXXXXX
------------------------------
46137555
SQL> select dbms_utility.data_block_address_file(46137555),
2 dbms_utility.data_block_address_block(46137555) from dual;
DBMS_UTILITY.DATA_BLOCK_ADDRES DBMS_UTILITY.DATA_BLOCK_ADDRES
------------------------------ ------------------------------
11 211
10bit的file number，每个对象最多有1022个文件（2个文件预留）
22bit的block number，每个文件最多有4M个BLOCK
及DBA中高10位是文件号，第22位是块号，一般我们使用dbms_utility进行转换，但是
有了算法就可以模拟一个换算的C语言小程序，利用C语言位域结构体
其实我们只要加权相加即可：
#include
#include
#include
#include
#define dn data->d
struct db_addr
{
unsigned d1:1;
unsigned d2:1;
unsigned d3:1;
unsigned d4:1;
unsigned d5:1;
unsigned d6:1;
unsigned d7:1;
unsigned d8:1;
unsigned d9:1;
unsigned d10:1;
unsigned d11:1;
unsigned d12:1;
unsigned d13:1;
unsigned d14:1;
unsigned d15:1;
unsigned d16:1;
unsigned d17:1;
unsigned d18:1;
unsigned d19:1;
unsigned d20:1;
unsigned d21:1;
unsigned d22:1;
unsigned d23:1;
unsigned d24:1;
unsigned d25:1;
unsigned d26:1;
unsigned d27:1;
unsigned d28:1;
unsigned d29:1;
unsigned d30:1;
unsigned d31:1;
unsigned d32:1;
};
void main ()
{
long dba,block1,block2,block3,block4,block5,file1,file2;
long blocksum,filesum;
struct db_addr *data;
printf("please input X dba!\n");
scanf("%x",&dba);
data=&dba;
block1=data->d1+data->d2*2+data->d3*pow(2,2)+data->d4*pow(2,3)+data->d5*pow(2,4);
// printf("%d",data->d1);
block2=data->d6*pow(2,5)+data->d7*pow(2,6)+data->d8*pow(2,7)+data->d9*pow(2,8)+data->d10*pow(2,9);
block3=data->d11*pow(2,10)+data->d12*pow(2,11)+data->d13*pow(2,12)+data->d14*pow(2,13)+data->d15*pow(2,14);
block4=data->d16*pow(2,15)+data->d17*pow(2,17)+data->d18*pow(2,17)+data->d19*pow(2,18)+data->d20*pow(2,19);
block5=data->d21*pow(2,20)+data->d22*pow(2,21);
blocksum=block1+block2+block3+block4+block5;
file1=data->d23*pow(2,0)+data->d24*pow(2,1)+data->d25*pow(2,2)+data->d26*pow(2,3)+data->d27*pow(2,4);
file2=data->d28*pow(2,5)+data->d29*pow(2,6)+data->d30*pow(2,7)+data->d31*pow(2,8)+data->d32*pow(2,9);
filesum=file1+file2;
printf("file id is:%ld\nblocks id is :%ld\n",filesum,blocksum);
}
如果我们使用程序换算
2c000d3
如下：
please input X dba!
2c000d3
file id is:11
blocks id is :211
实际和dbms_utility得到的是相同的。
SQL> select to_number('2c000d3','xxxxxxxxxxxxxx') from dual;
TO_NUMBER('2C000D3','XXXXXXXXX
------------------------------
46137555
SQL> select dbms_utility.data_block_address_file(46137555),
2 dbms_utility.data_block_address_block(46137555) from dual;
DBMS_UTILITY.DATA_BLOCK_ADDRES DBMS_UTILITY.DATA_BLOCK_ADDRES
------------------------------ ------------------------------
11 211
