看上图，debugfs出来的信息，Group 2，并没有所谓的组描述符。而Group1，用8194和8195两个块来存储。OK，我们看下，里面存储的是什么东西。
Group 0里面第2和第3块存储的是组描述符，也就说从0x000800～0x001000是组描述符块的内容。
- 000800 04 01 00 00 05 01 00 00 06 01 00 00 ef 1d e5 07
- 000810 02 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00 块组0的组描述符
- -----------------------------------------------------------------------
- 000820 04 21 00 00 05 21 00 00 06 21 00 00 fd 1d f0 07
- 000830 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00 块组1的组描述符
- -----------------------------------------------------------------------
- 000840 01 40 00 00 02 40 00 00 03 40 00 00 00 1f f0 07
- 000850 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00 块组2的组描述符
- ------------------------------------------------------------------------
- 000860 04 61 00 00 05 61 00 00 06 61 00 00 fd 1d f0 07
- 000870 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000880 01 80 00 00 02 80 00 00 03 80 00 00 00 1f f0 07
- 000890 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 0008a0 04 a1 00 00 05 a1 00 00 06 a1 00 00 fd 1d f0 07
- 0008b0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 0008c0 01 c0 00 00 02 c0 00 00 03 c0 00 00 00 1f f0 07
- 0008d0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 0008e0 04 e1 00 00 05 e1 00 00 06 e1 00 00 fd 1d f0 07
- 0008f0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000900 01 00 01 00 02 00 01 00 03 00 01 00 00 1f f0 07
- 000910 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000fb0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000fc0 01 c0 07 00 02 c0 07 00 03 c0 07 00 ff 0e f0 07
- 000fd0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00 块组62的组描述符
- -----------------------------------------------------------------------
- 000fe0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
-
- * 没有块组63
- ----------------------------------------------------------------------
- 001000 04 20 00 00 04 60 00 00 04 a0 00 00 04 e0 00 00
-
04 01 00 00 转换成可读的十进制是0x104=259，表示数据位图位于第259块block。inode位图位于260，和debugfs出来的信息是一样的（不算启动块）。0x1def=7663个空闲数据块....
各位看官可以自己解析任何一个块组的相关信息，可以证明和debugfs出来的块组的信息是一致的。现在我们确定了，组描述符以数组的形式存储在K个快上，对于我们只有63个组块，每个组块需要32个字节，只需要2个1KB的block就足够了。这就是说，其实组描述符和超级块一样，其实是冗余的。也就是说，其他存储组描述符的两个block，信息和块组0中的组描述符的两个block是一样的。下面我来证明。
块组25也有组描述符块，204802和204803两个块，记录了63个块组的组描述符信息。内容应该和前面的块组0的两个块一致。我已经取出了这两个block的内容，大家自己比较吧，结果是内容是一样的。
- Group 25: (Blocks 204801-212992)
- 备份 superblock at 204801, Group descriptors at 204802-204803
- 保留的GDT块位于 204804-205059
- Block bitmap at 205060 (+259), Inode bitmap at 205061 (+260)
- Inode表位于 205062-205315 (+261)
- 7677 free blocks, 2032 free inodes, 0 directories
- 可用块数: 205316-212992
- 可用inode数: 50801-52832
点击(此处)折叠或打开
- root@libin:/mnt/bean# dd if=/dev/loop0 bs=1k skip=204802 count=2|od -tx1 -Ax > /tmp/dump_hex_
- 记录了2+0 的读入
- 记录了2+0 的写出
- 2048字节(2.0 kB)已复制，0.000160205 秒，12.8 MB/秒
- root@libin:/mnt/bean# vi /tmp/dump_hex_
-
-
- 000000 04 01 00 00 05 01 00 00 06 01 00 00 ef 1d e5 07
- 000010 02 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000020 04 21 00 00 05 21 00 00 06 21 00 00 fd 1d f0 07
- 000030 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000040 01 40 00 00 02 40 00 00 03 40 00 00 00 1f f0 07
- 000050 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000060 04 61 00 00 05 61 00 00 06 61 00 00 fd 1d f0 07
- 000070 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 000080 01 80 00 00 02 80 00 00 03 80 00 00 00 1f f0 07
- 000090 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 0000a0 04 a1 00 00 05 a1 00 00 06 a1 00 00 fd 1d f0 07
- 0000b0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- ....
-
- 0007c0 01 c0 07 00 02 c0 07 00 03 c0 07 00 ff 0e f0 07
- 0007d0 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00
- 0007e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
- *
- 000800
最后，最后的最后，解释以下，为什么每个块组中的块数blocks per group 是8192,因为，我们用1个块作为位图保存本块组 block的使用情况（bit为1表示对应的block被使用，bit为0表示对应的block空闲），1个block是1024字节，共有1024*8=8192个bit，所以，每个块组最多只能是81292个块。
同样道理如果用户使用的是4094大小的块，那么，4096*8=32768个bit，所以每个块组会有32K个块。证据在下面。 |
