解码一个JPEG霍夫曼块（表） [英] Decoding a JPEG Huffman block (table)
问题描述
以下块由霍夫曼块标记嵌套
The following block is nested by Huffman block markers
-HUFF---------------------------------------------------------------------0084-
10 0 1 2 4 3 4 6 5 6 8 a 9 4 2 3
0 1 2 11 0 3 4 21 5 12 31 6 41 51 61 13
22 71 81 91 a1 14 32 b1 d1 f0 15 23 35 42 b2 c1
7 16 24 33 52 72 73 e1 25 34 43 53 62 74 82 94
a2 f1 26 44 54 63 64 92 93 c2 d2 55 56 84 b3 45
83 46 a3 e2
-------------------------------------------------------------------------------
0084是表作为整数的长度，不包含在该块此处
0084 is the length of the table as an integer and is not included in the block here
，第一地址aparently使得它的AC表在目的地0（0×10）
according to the JPEG standard, the first address aparently makes it an AC table at destination 0 (0x10)
和aparently从那里开始它是一个哈夫曼表。
and aparently from there onwards it's a huffman table.
那么，它是如何去codeD？
So, how is it decoded?
推荐答案
后，接下来的16字节为0x10告诉你每个长度有多少codeS。在你的榜样，有长度为1位的0 codeS，长度为2位的$ C $ 1℃，2 $ C $长度为3位的CS，4 codeS长度为4位，3 $ C $长度为5比特的CS，等
The next 16 bytes after the 0x10 tell you how many codes of each length. In your example, there are 0 codes of length 1 bit, 1 code of length 2 bits, 2 codes of length 3 bits, 4 codes of length 4 bits, 3 codes of length 5 bits, and so on.
这些都是再接着是连接这些codeS codeD，以便值。再从你的例子：
These are then followed by the values that are encoded by those codes, in order. Again from your example:
Code length | Number | Symbol(s)
------------+--------+----------
1 bit | 0 |
2 bits | 1 | 0x01
3 bits | 2 | 0x02 0x11
4 bits | 4 | 0x00 0x03 0x04 0x21
5 bits | 3 | 0x05 0x12 0x31
... etc
您然后建立自上而下二叉树，为了分配的符号。在这个例子中，您可以：
You then build a binary tree from the top down, assigning the symbols in order. In this example, you get:
Symbol | Code
-------+------
0x01 | 00
0x02 | 010
0x11 | 011
0x00 | 1000
0x03 | 1001
0x04 | 1010
0x21 | 1011
...etc
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