oracle如何识别 “脏块”

你好Oracle大师，
我有一个非常基本的问题来了解oracle的内部行为。我读了几个博客，不幸的是最终陷入了混乱。所以想问你 ..

假设，我正在升级一个区块 (记录)，但没有发布 “委员会”。

发生检查点时，数据库编写器将脏块写入磁盘。

问题:
假设当另一个用户查询同一块 (未提交) 时，oracle如何知道 “该” 特定块是脏的，并且应该从 “撤消” 日志中读取 (以保持一致性)？
比如，oracle是否在块标头中维护标志？

它在块标题上。你可以从v $ bh看到这个，

SQL> alter system flush buffer_cache;

System altered.

SQL> create table t ( x primary key, y ) as select 1 x, 2 y from dual;

Table created.

SQL> select data_object_id from user_objects where object_name = 'T';

DATA_OBJECT_ID
--------------
        101919

SQL>
SQL> select
  2    dbms_rowid.rowid_relative_fno(rowid) f,
  3    dbms_rowid.rowid_block_number(rowid) b
  4  from t;

         F          B
---------- ----------
         7      16259

--
-- here's the freshly created block
--

SQL> select status
  2  from   v$bh
  3  where  file# = 7
  4  and    block# =  16259
  5  and    objd = 101919
  6  /

STATUS
----------
xcur

--
-- if I flush the buffer cache, the block is still "there" but its marked as free, ie, I could replace it
-- with something else
--
SQL> alter system flush buffer_cache;

System altered.

SQL> select status
  2  from   v$bh
  3  where  file# = 7
  4  and    block# =  16259
  5  and    objd = 101919
  6  /

STATUS
----------
free

--
-- I query it back in, and it is the current version of this block because
-- nothing else has ever need to read it consistently
--
SQL> select * from t where x=1;

         X          Y
---------- ----------
         1          2

SQL> select objd, status
  2  from   v$bh
  3  where  file# = 7
  4  and    block# =  16259
  5  and    objd = 101919
  6  /

      OBJD STATUS
---------- ----------
    101919 xcur

--
-- now I delete it (without commit)
--
SQL> delete from t;

SQL> select objd, status, dirty
  2  from   v$bh
  3  where  file# = 7
  4  and    block# =  16259
  5  and    objd = 101919
  6  /

      OBJD STATUS     DIRTY
---------- ---------- -----
    101919 xcur           Y

--
-- and now another session gets an older version of it via a query
--
SQL> select * from t where x=1;

         X          Y
---------- ----------
         1          2

--
-- and we can see the consistent version be created
--

SQL> select objd, status
  2  from   v$bh
  3  where  file# = 7
  4  and    block# =  16259
  5  and    objd = 101919
  6  /

      OBJD STATUS
---------- ----------
    101919 xcur
    101919 cr

但是在最简单的概念形式中，SCN可以驱动所有这些。

当我运行查询时，我实际上是在说 “我需要所有块作为SCN 1234” (即，当查询开始时SCN)。

当我遇到一个块时，无论它是否脏，是否已提交，我的要求将是-该块上的SCN是否等于或在1234之前。如果不是，那么这个块对我来说太 “新” 了，这是一个信号，表明我需要得到它的旧版本。