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标题: 深入解析 ext2 文件系统（4） [打印本页]

作者: yuyang911220 时间: 2016-9-18 23:01 标题: 深入解析 ext2 文件系统（4）

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。

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