问题描述
这是有关bzip2 的问题。任何Bzip2归档文件都由文件头，一个或多个块和尾部结构组成。所有块均应以Pi编号为0x314159265359的6个字节的BCD编码数字 1AY& SY开头。根据：
This is the question about bzip2 archive format. Any Bzip2 archive consists of file header, one or more blocks and tail structure. All blocks should start with "1AY&SY", 6 bytes of BCD-encoded digits of the Pi number, 0x314159265359. According to the source of bzip2:
/*--
A 6-byte block header, the value chosen arbitrarily
as 0x314159265359 :-). A 32 bit value does not really
give a strong enough guarantee that the value will not
appear by chance in the compressed datastream. Worst-case
probability of this event, for a 900k block, is about
2.0e-3 for 32 bits, 1.0e-5 for 40 bits and 4.0e-8 for 48 bits.
For a compressed file of size 100Gb -- about 100000 blocks --
only a 48-bit marker will do. NB: normal compression/
decompression do *not* rely on these statistical properties.
They are only important when trying to recover blocks from
damaged files.
--*/
问题是：确实如此，所有bzip2存档都会有与字节边界开始对齐的块？我的意思是所有由bzip2的引用实现（bzip2-1.0.5 +实用程序）创建的所有档案。
The question is: Is it true, that all bzip2 archives will have blocks with start aligned to byte boundary? I mean all archives created by reference implementation of bzip2, the bzip2-1.0.5+ utility.
我认为bzip2可能不是将字节流解析为字节流，而是将位解析为位流（该块本身由霍夫曼编码，在设计上未按字节对齐）。
I think that bzip2 may parse the stream not as byte stream but as bit stream (the block itself is encoded by huffman, which is not byte-aligned by design).
因此，换句话说：如果 grep -c 1AY& SY 更大（霍夫曼内部可能会产生1AY& SY
So, in other words: If grep -c 1AY&SY greater (huffman may generate 1AY&SY inside block) or equal to count of bzip2 blocks in the file?
推荐答案
BZIP2会查看比特流。
BZIP2 looks at a bit stream.
来自：
bzip2 艺术条款还暗示了位块对齐（请参阅文件格式部分），这似乎与我在学校时记得的内联（必须实现该算法...）。
The bzip2 wikipedia article also alludes to bit-block alignment (see the File Format section), which seems to be inline from what I remember from school (had to implement the algorithm...).
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