c jpeg 理论霍夫曼 DC AC表，c程序实现正向逆向转换

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此4张表是理论表，不是针对某张图片的特定表。如编码程序不统计生成某图片的专用霍夫曼表，应该也可用理论表代用编码。

1.亮度DC表

左边第一列是二进制位数，就是对此位数编码

中间一列是生成比特流的位数，右边是生成的比特流。

2.色度DC表

3.亮度AC表

4.  色度AC表

 利用这4张表转换生成亮度，色度的DC，AC比特流

图3图4两张表是用jpeg文件头的形式显示的。

AC 表被编码的数值是两部分组成的，高4比特是0的个数 （a,b）对中的a，低4节为b的二进制位数。

两个AC表中的162个数就是两个输入参数组成的一个字节的值。

对于DC表，直接用if一一例举出现就行了。AC两张表输出都是162个数，这162个数的比特流从1位到16位都有。

  看了AC两张表，最多的是16位的比特流。

AC表的输入为0的个数和二进制位数两个参数，输出为比特流和比特流宽度。

--------------------------------

亮度AC表：

2 位：2个     00(1) ，01(2)

3位：1      010(0),

4位：3      0110(4)，0111(11), 1000(5)

5位：3     10010(12) ,10011(21),10100(31)

6位：2     101010(41), 101011(6)

7位：4      1011000(13),1011001(51), 1011010(61),1011011(7)

8位：3        10111000(22),  10111001(71),  10111010(14)

9位：5        101110110(32),  101110111(81), 101111000(91)  ，101111001(A1), 101111010(8)

10位：5      1011110110(23),           1011110111(42),1011111000(b1), 1011111001(c1),1011111010(15)

11位：4     10111110110(52),   10111110111(d1),  10111111000(f0), 10111111001(24)

12位：4     101111110100(33),   101111110101(62), 101111110110(72),101111110111(82)

15位：1     101111111000000(9),

16位 ：125     1011111110000010(0a),，，，，，，

共计162个

亮度AC转换：本想用查表法代码太长了，162个if语句。现还不包括16位的数据转换，准备换一种方法。DC可以用这种方法。

​1. 亮度AC编码用正向程序

​
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/ioctl.h>
#include <linux/videodev2.h>  
#include <string.h>
#include <sys/mman.h>
#include <linux/fb.h>
#include <math.h>int main(void) {unsigned char i_0=0xf;unsigned char i_len=0xa;unsigned char len;unsigned int o;unsigned char zj=i_0*16+i_len;      //合成一个字节unsigned char  ws[16]={0,2,1,3,3,2,4,3,5,5,4,4,0,0,1,0x7d};unsigned char zh[162]={0x1, 0x2, 0x3, 0x0, 0x4, 0x11, 0x5,0x12,0x21,0x31,0x41,0x6, 0x13,0x51,0x61,0x7,0x22,0x71,0x14,0x32,0x81,0x91,0xa1,0x08,0x23,0x42,0xb1,0xc1,0x15,0x52,0xd1,0xf0,0x24,0x33,0x62,0x72,0x82,0x09,0x0a,0x16,0x17,0x18,0x19,0x1a,0x25,0x26,0x27,0x28,0x29,0x2a,0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a,0x83,0x84,0x85,0x86,0x87,0x88,0x89,0x8a,0x92,0x93,0x94,0x95,0x96,0x97,0x98,0x99,0x9a,0xa2,0xa3,0xa4,0xa5,0xa6,0xa7,0xa8,0xa9,0xaa,0xb2,0xb3,0xb4,0xb5,0xb6,0xb7,0xb8,0xb9,0xba,0xc2,0xc3,0xc4,0xc5,0xc6,0xc7,0xc8,0xc9,0xca,0xd2,0xd3,0xd4,0xd5,0xd6,0xd7,0xd8,0xd9,0xda,0xe1,0xe2,0xe3,0xe4,0xe5,0xe6,0xe6,0xe7,0xe8,0xe9,0xea,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf9,0xfa};int cx_ws,cx_b;unsigned char hfm[17][0x7d]={};int t=0;for(int a=0;a<16;a++){             //把要编码的162个数按位数分为16个数组，二位一组....16位一组if(ws[a]==0){continue;}for(int b=0;b<ws[a];b++){hfm[a+1][b]=zh[t];t++;}}for(int a=0;a<16;a++){               //查询输入数的位数和在所在组的顺序，组内顺序从0开始if(ws[a]==0){continue;}for(int b=0;b<ws[a];b++){if(	hfm[a+1][b]==zj){cx_ws=a+1;         //得到输入数二进制位数，根据此数到相依位数的数组查询cx_b=b;            //输入数在按位数分组的数组中所在的位置，从0开始break;}}}int o_js=0;       //每一组的第一个数的值if(cx_ws==2){o_js=0;}if(cx_ws==3){o_js=0b010;}if(cx_ws==4){o_js=0b0110;}if(cx_ws==5){o_js=0b10010;}if(cx_ws==6){o_js=0b101010;}if(cx_ws==7){o_js=0b1011000;}if(cx_ws==8){o_js=0b10111000;}if(cx_ws==9){o_js=0b101110110;}if(cx_ws==10){o_js=0b1011110110;}if(cx_ws==11){o_js=0b10111110110;}if(cx_ws==12){o_js=0b101111110100;}if(cx_ws==15){o_js=0b101111111000000;}if(cx_ws==16){o_js=0b1011111110000010;}len=cx_ws;o=o_js+cx_b;printf("len:%d o:%d\n",len,o);return 0;	
}​​

2.亮度AC 解码程序

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/ioctl.h>
#include <linux/videodev2.h>  
#include <string.h>
#include <sys/mman.h>
#include <linux/fb.h>
#include <math.h>int main(void) {unsigned int i_bit=0b1011111110000011;unsigned char i_len=16;unsigned char len;unsigned char o;unsigned char  ws[16]={0,2,1,3,3,2,4,3,5,5,4,4,0,0,1,0x7d};unsigned char zh[162]={0x1, 0x2, 0x3, 0x0, 0x4, 0x11, 0x5,0x12,0x21,0x31,0x41,0x6, 0x13,0x51,0x61,0x7,0x22,0x71,0x14,0x32,0x81,0x91,0xa1,0x08,0x23,0x42,0xb1,0xc1,0x15,0x52,0xd1,0xf0,0x24,0x33,0x62,0x72,0x82,0x09,0x0a,0x16,0x17,0x18,0x19,0x1a,0x25,0x26,0x27,0x28,0x29,0x2a,0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a,0x83,0x84,0x85,0x86,0x87,0x88,0x89,0x8a,0x92,0x93,0x94,0x95,0x96,0x97,0x98,0x99,0x9a,0xa2,0xa3,0xa4,0xa5,0xa6,0xa7,0xa8,0xa9,0xaa,0xb2,0xb3,0xb4,0xb5,0xb6,0xb7,0xb8,0xb9,0xba,0xc2,0xc3,0xc4,0xc5,0xc6,0xc7,0xc8,0xc9,0xca,0xd2,0xd3,0xd4,0xd5,0xd6,0xd7,0xd8,0xd9,0xda,0xe1,0xe2,0xe3,0xe4,0xe5,0xe6,0xe6,0xe7,0xe8,0xe9,0xea,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf9,0xfa};unsigned char  cx_ws,cx_b;unsigned char hfm[17][0x7d]={};int t=0;for(int a=0;a<16;a++){             if(ws[a]==0){continue;}for(int b=0;b<ws[a];b++){hfm[a+1][b]=zh[t];t++;}}
//---------------------------------------cx_ws=i_len;int o_js=0;      if(cx_ws==2){o_js=0;cx_b=i_bit-o_js;}if(cx_ws==3){o_js=0b010;cx_b=i_bit-o_js;}if(cx_ws==4){o_js=0b0110;cx_b=i_bit-o_js;}if(cx_ws==5){o_js=0b10010;cx_b=i_bit-o_js;}if(cx_ws==6){o_js=0b101010;cx_b=i_bit-o_js;	}if(cx_ws==7){o_js=0b1011000;cx_b=i_bit-o_js;}if(cx_ws==8){o_js=0b10111000;cx_b=i_bit-o_js;}if(cx_ws==9){o_js=0b101110110;cx_b=i_bit-o_js;}if(cx_ws==10){o_js=0b1011110110;cx_b=i_bit-o_js;}if(cx_ws==11){o_js=0b10111110110;cx_b=i_bit-o_js;}if(cx_ws==12){o_js=0b101111110100;cx_b=i_bit-o_js;}if(cx_ws==15){o_js=0b101111111000000;cx_b=i_bit-o_js;}if(cx_ws==16){o_js=0b1011111110000010;cx_b=i_bit-o_js;}//-----------------------------------------unsigned char o_zj=hfm[cx_ws][cx_b];len=o_zj/16;o=o_zj%16;printf("%x  %x",len,o);return 0;	
}

 

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