H.264中整数DCT变换，量化，反量化，反DCT究竟是如何实现的？(无代码，无真相)

来源：互联网发布：淘宝网买东西的步骤编辑：程序博客网时间：2024/06/05 22:41

H.264中采用的是整数DCT变换，在实现的时候，该变换和量化又杂糅在一起，那么这些错综复杂的关系究竟是怎样纠缠的呢？在参考H.264乐园论坛会员cs1860wd的帖子和H.264 and MPEG-4 VIDEO COMPRESSION(第一版)这本书后，基于帖子和书上的讲解，给出相应的实现代码，并验证代码的正确性.

还是以foreman视频第一帧第一个宏块第一个4*4块为例. 下面给出像素值：

====================== Y Data ======================
+----------------+----------------+----------------+----------------+
| 43,216,254,249,|251,254,254,253,|251,252,254,254,|254,254,254,253,|
| 49,198,193,211,|228,205,213,185,|211,207,186,248,|198,203,208,183,|
| 48,194,177,171,|197,173,185,136,|191,195,138,179,|142,176,177,135,|
| 46,214,225,169,|177,189,198,160,|203,208,177,165,|173,196,191,156,|
+----------------+----------------+----------------+----------------+
| 41,185,208,180,|203,228,226,200,|214,226,225,227,|228,225,224,210,|
| 31,130,173,178,|215,230,221,212,|220,229,227,228,|229,227,226,226,|
| 29,119,194,216,|211,213,219,222,|225,223,220,219,|218,218,218,218,|
| 25,126,219,224,|217,224,227,227,|227,226,225,224,|220,220,221,222,|
+----------------+----------------+----------------+----------------+
| 26,131,215,223,|226,225,225,225,|225,226,223,219,|221,221,219,220,|
| 30,136,216,226,|223,224,225,225,|224,221,217,221,|222,219,220,226,|
| 30,136,216,227,|224,224,225,223,|221,218,221,216,|211,224,224,211,|
| 29,135,217,225,|222,221,222,222,|221,209,181,155,|186,210,186,164,|
+----------------+----------------+----------------+----------------+
| 29,134,216,224,|226,230,230,227,|206,177,146,113,|149,162,147,150,|
| 29,135,219,231,|225,201,190,185,|163,144,153,140,|127,143,165,184,|
| 30,139,210,192,|165,142,134,133,|143,141,129,138,|150,178,201,207,|
| 30,125,166,145,|144,154,132,111,|118,161,175,180,|204,214,213,209,|
+----------------+----------------+----------------+----------------+

该块的预测值为128（16个位置都是128），之前分析过，在JM8.6中，这一块在编码端和解码端的QDCT均为：

9 -12 -11 -5

3 -3 1 0

3 -1 -2 1

0 0 0 0

且用H.264visa从码流中得到的DCT系数为：（注意解码端的DCT与编码端的DCT必然不同）

====================== Y Data ======================
+------------------------+------------------------+------------------------+------------------------+
| 2304,-3840,-2816,-1600,| 768, 640, -256, 640,| 1280, 320, 256, -640,| 768, -320, -768,    0,|
| 960,-1200, 320, 0,|    0,    0, 320,    0,| -640, -800,    0,    0,| 960, -800, 320,    0,|
| 768, -320, -512, 320,| 512, -640,    0, -320,| -768, 320, -512, 320,| 768,    0, 256,    0,|
|    0,    0,    0,    0,|    0,    0,    0,    0,|    0, 400,    0,    0,| -320,    0,    0,    0,|
+------------------------+------------------------+------------------------+------------------------+
|-1024,-5120,-1792, -640,| 2560, -640, -256,    0,| 512,    0,    0,    0,|    0,    0,    0,    0,|
|    0, 1200, -640, -400,| -320, 400,    0,    0,| 640,    0,    0,    0,| 320,    0,    0,    0,|
| 512,    0, -256,    0,|    0,    0,    0,    0,|    0,    0,    0,    0,|    0,    0,    0,    0,|
| 320,    0,    0,    0,| -320,    0,    0,    0,|    0,    0,    0,    0,|    0,    0,    0,    0,|
+------------------------+------------------------+------------------------+------------------------+
|    0, 320,    0,    0,|    0,    0,    0,    0,| -768, 640,    0,    0,| 512,    0, -256,    0,|
|    0,    0,    0,    0,|    0,    0,    0,    0,| 640, -800,    0,    0,| -640, -400, 320,    0,|
|    0,    0,    0,    0,|    0,    0,    0,    0,| -256, 320,    0,    0,|    0, 320,    0,    0,|
|    0,    0,    0,    0,|    0,    0,    0,    0,| 320, -400,    0,    0,| -320,    0,    0,    0,|
+------------------------+------------------------+------------------------+------------------------+
| -512, 640, -256,    0,|    0,    0,    0,    0,| 1024, -320, -256,    0,| 1024, -320,    0,    0,|
| 960,-1200, 320,    0,| 960, -800, 320,    0,|-1280, 800,    0, 400,| -640, 400,    0,    0,|
| -512, 320,    0,    0,|    0,    0, -256,    0,|    0,    0, -256,    0,| 512, 640,    0,    0,|
|    0,    0,    0,    0,| -320,    0,    0,    0,| -640, 800,    0,    0,|    0,    0,    0,    0,|
+------------------------+------------------------+------------------------+------------------------+

下面根据变换，量化，反量化和反变换公式给出如下程序，看看是否与上述数据一致.(需要特别指出的是：JM8.6中并不是这么实现的，但本质是相同的，最后结果也一样.)

#include <iostream>#include <cmath>#define BLOCK_SIZE 4using namespace std;int QP = 28; // 量化参数//原始的YUV矩阵int orgYUV[BLOCK_SIZE][BLOCK_SIZE] ={43, 216, 254, 249,    49, 198, 193, 211,    48, 194, 177, 171,    46, 214, 225, 169};//预测值矩阵int predYUV[BLOCK_SIZE][BLOCK_SIZE] ={128, 128, 128, 128,128, 128, 128, 128,128, 128, 128, 128,128, 128, 128, 128};int D[BLOCK_SIZE][BLOCK_SIZE];  //中间矩阵int Di[BLOCK_SIZE][BLOCK_SIZE]; //中间矩阵int W[BLOCK_SIZE][BLOCK_SIZE];  //核矩阵int Z[BLOCK_SIZE][BLOCK_SIZE];  //QDCT矩阵int Wi[BLOCK_SIZE][BLOCK_SIZE]; //Wi矩阵int Xi[BLOCK_SIZE][BLOCK_SIZE]; //解码的残差矩阵//Cf矩阵int Cf[BLOCK_SIZE][BLOCK_SIZE]={    1, 1,  1, 1,    2, 1, -1, -2,    1,-1, -1, 1,    1,-2,  2, -1};//Ci矩阵      int Ci[BLOCK_SIZE][BLOCK_SIZE]={    2,  2,  2,  1,    2,  1, -2, -2,    2, -1, -2,  2,    2, -2,  2, -1};//MF矩阵int MF[6][3]={    13107, 5243, 8066,    11916, 4660, 7490,    10082, 4194, 6554,    9362,  3647, 5825,    8192,  3355, 5243,    7282,  2893, 4559};//Qstep矩阵int V[6][3]={    10, 16, 13,    11, 18, 14,    13, 20, 16,    14, 23, 18,    16, 25, 20,    18, 29, 23};//矩阵转置void matrixTransform(int a[][BLOCK_SIZE]){int i, j, tmp;for(i = 0; i < BLOCK_SIZE; i++){for(j = 0; j < BLOCK_SIZE; j++){if(i < j){tmp = a[i][j];a[i][j] = a[j][i];a[j][i] = tmp;}}}}//矩阵求差void  matrixSubtract(int a[][BLOCK_SIZE],int b[][BLOCK_SIZE]){    int i, j;    for(i = 0;i < BLOCK_SIZE; i++){        for(j = 0; j < BLOCK_SIZE; j++)        {            a[i][j] -= b[i][j];}}}//矩阵求积void  matrixMultiply(int a[][BLOCK_SIZE],int b[][BLOCK_SIZE],int c[][BLOCK_SIZE]){int i, j, k;for(i = 0; i < BLOCK_SIZE; i++){for(j = 0; j < BLOCK_SIZE; j++){c[i][j] = 0;for(k = 0; k < BLOCK_SIZE; k++){c[i][j] += a[i][k] * b[k][j];  }}}      }//矩阵显示void matrixShow(int a[][BLOCK_SIZE]){    int i, j;    cout << "*****************************" << endl;    for(i = 0; i < BLOCK_SIZE; i++){        for(j = 0; j < BLOCK_SIZE; j++)        {            cout << a[i][j] << "\t";        }cout << endl;}    cout << "*****************************" << endl << endl;}//求QDCTvoid quantizeDCT(int W[][BLOCK_SIZE]){// QP决定了qbits和f, QP和位置(i, j)共同决定了mfint qbits = 15 + floor(QP / 6);int f = (int)( pow(2.0, qbits) / 3 );int mf;int i, j, k;for(i = 0; i < BLOCK_SIZE; i++){for(j = 0; j < BLOCK_SIZE; j++){//以下均依公式实现//(0, 0), (2, 0), (0, 2), (2, 2)  mf为 MF[QP % 6][0]//(1, 1), (3, 1), (1, 3), (3, 3)  mf为 MF[QP % 6][1];// other positions                mf为 MF[QP % 6][2];if((0 == i || 2 == i) && (0 == j || 2 == j))k = 0;else if((1 == i || 3 == i) && (1 == j || 3 == j))k = 1;elsek = 2;mf = MF[QP % 6][k];              Z[i][j] = ( abs(W[i][j]) * mf + f ) >> qbits;if(W[i][j] < 0)Z[i][j] = -Z[i][j];} }}//求Wi(即解码端的DCT) (Z为QDCT)void reverseQuantize(int Z[][BLOCK_SIZE]){     int t = floor(QP / 6);                int f = (int)pow(2, t); int v;int i, j, k;for(i = 0; i < BLOCK_SIZE; i++){for(j = 0; j < BLOCK_SIZE; j++){//以下均依公式实现  if((0 == i || 2 == i) && (0 == j || 2 == j))k = 0;else if((1 == i || 3 == i) && (1 == j || 3 == j))k = 1;elsek = 2;v = V[QP % 6 ][k];Wi[i][j] = Z[i][j] * v * f;}}}int main(){matrixSubtract(orgYUV, predYUV);cout << "Residual matrix is" << endl;    matrixShow(orgYUV);//此时orgYUV变为残差矩阵    matrixMultiply(Cf, orgYUV, D);    matrixTransform(Cf);    matrixMultiply(D, Cf, W);        //得到的W即为核cout << "Matrix Core（W） is" << endl;    matrixShow(W);//利用核W来得到QDCT（Z）    quantizeDCT(W);cout << "Matrix QDCT(Z) is" << endl;    matrixShow(Z);      //利用QDCT(Z)得到解码端的DCT(Wi).(Wi与编码端DCT必然不同）    reverseQuantize(Z);cout << "Matrix W'(解码端DCT) is" << endl;    matrixShow(Wi);    //利用Wi得到解码的残差矩阵Xi    matrixMultiply(Ci, Wi, Di); matrixTransform(Ci);            matrixMultiply(Di, Ci, Xi);int i,j;    for(i = 0;i < 4; i++){        for(j = 0; j < 4; j++){Xi[i][j] = int( Xi[i][j] / 256.0 + 0.5 );}}cout << "Matrix Xi(解码端残差) is" << endl;    matrixShow(Xi);return 0;}

结果为：(可以看出，结果中的QDCT和解码端的DCT都与之前的数据吻合，从而证明上面程序的实现是正确的.)

Residual matrix is
*****************************
-85     88      126     121
-79     70      65      83
-80     66      49      43
-82     86      97      41
*****************************

Matrix Core（W） is
*****************************
609     -1255   -685    -560
277     -476    113     -73
175     -159    -119    98
-14     -13     4       1
*****************************

Matrix QDCT(Z) is
*****************************
9       -12     -11     -5
3       -3      1       0
3       -1      -2      1
0       0       0       0
*****************************

Matrix W'(解码端DCT) is
*****************************
2304    -3840   -2816   -1600
960     -1200   320     0
768     -320    -512    320
0       0       0       0
*****************************

Matrix Xi(解码端残差) is
*****************************
-77     88      132     110
-80     63      67      77
-82     62      48      39
-79     87      93      32
*****************************

下面，简要看看整数DCT变换的原理.(关于整数DCT变换公式的推倒，请参考余兆明的《图像编码标准H.264技术》).

DCT变换为: DCT = A * X * At

量化为: QDCT = floor(DCT/Qstep)

反量化为: DCT‘ = QDCT * Qstep

反DCT为: X’ = At * DCT‘ * A

但在H.264中采用的是整数DCT, 在JM8.6中的实现方式也很有讲究：( .* 表示矩阵对应元素相乘)

整数DCT变化为： DCT = (Cf * X *Cft) .*Ef

量化为： QDCT = floor(DCT/Qstep)

反量化为： DCT‘ = QDCT * Qstep

整数反DCT变换为： X’ = Ci * (DCT‘ .* Ei) *Ci

在实现的时候，经常不直接得到DCT系数，而是将变换和量化结合在一起实现. 整数DCT变换有很多好处：没有除法，没有浮点数，所以高效且准确，而且减少了矩阵运算.

最后感慨一下：如果上面的程序用matlab来仿真，就简单多了，matlab太适合处理矩阵问题了.

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