Oracle raw date form

Question

Write a SQL statement that will obtain a date in Oracle in its raw form. And display the 4-digit year.

Answer 1

The 8 byte version comes from a C programming language structure, used in the Oracle kernel. It is similar to the standard tm structure (sorry, not the time_t as I said earlier) and works like this.
Byte 1: Epoch Number
Byte 2: Year offset from epoch base
Byte 3: Month
Byte 4: Day
Byte 5: Hour (24 hour time)
Byte 6: Minute
Byte 7: Second
Byte 8: I have no idea
All of the values in the structure appear to be unsigned byte type, so The base (byte 1) for the year offset (byte 2) needs to change, every 255 years.
SQL> select sysdate, dump(sysdate) from dual;
SYSDATE              DUMP(SYSDATE)
21-jul-2011 16:58:35 Typ=13 Len=8: 7,219,7,21,16,58,35,0
SQL> select 2011 - 219 from dual;
  2010-218
      1792
SQL> select dump(to_date('21-jul-1790', 'dd-mon-yyyy')) from dual;
DUMP(TO_DATE('21-JUL-1790','DD-M
Typ=13 Len=8: 6,254,7,21,0,0,0,0
SQL> select dump(to_date('21-jul-1792', 'dd-mon-yyyy')) from dual;
DUMP(TO_DATE('21-JUL-1792','DD
Typ=13 Len=8: 7,0,7,21,0,0,0,0john