這個2D DCT代碼是如何工作的？

Question

下面是代碼：

void dct(const tga_image *tga, double data[8][8], 
    const int xpos, const int ypos) 
{ 
    int i,j; 
    double in[8], out[8], rows[8][8]; 

    /* transform rows */ 
    for (j=0; j<8; j++) 
    { 
     for (i=0; i<8; i++) 
      in[i] = (double) pixel(tga, xpos+i, ypos+j); 
     dct_1d(in, out, 8); 
     for (i=0; i<8; i++) rows[j][i] = out[i]; 
    } 

    /* transform columns */ 
    for (j=0; j<8; j++) 
    { 
     for (i=0; i<8; i++) 
      in[i] = rows[i][j]; 
     dct_1d(in, out, 8); 
     for (i=0; i<8; i++) data[i][j] = out[i]; 
    } 
} 

它是從https://unix4lyfe.org/dct/

我只有1個問題找到listing2.c拍攝，我們填寫的行，行[J] [I]，但隨後讀出來out rows [i] [j]。根據2D DCT公式，我們轉置DCT矩陣而不是實際數據。爲什麼實際數據正在轉換？

Answer 1

如果我假設xpos爲水平索引而ypos爲垂直索引，則以下情況屬實。

dct_1d(*,*,*);適用於僅1-d陣列（這裏in和out）中的作用。你得到的結果是indexing笨拙二維陣列在C在這裏（特別是rows這裏）。

void dct(const tga_image *tga, double data[8][8], 
    const int xpos, const int ypos) 
{ 
    int i,j; /* as in matrix[i][j] */ 
    double in[8], out[8], rows[8][8]; 

    /* transform rows (each is running horizontally with j) */ 
    for (i=0; i<8; i++) 
    { 
     for (j=0; j<8; j++) 
      in[j] = (double) pixel(tga, xpos+j, ypos+i); /* fill current row i */ 
     /* Note above xpos in an image is horizontal as j is in a matrix[i][j] in c and 
     vice versa. (The fallacy that you will make is the following: You will think that 
     xpos corresponds to i and ypos corresponds to j, which is incorrect.) */ 
     dct_1d(in, out, 8); /* transform current row i */ 
     for (j=0; j<8; j++) rows[i][j] = out[j]; /* copy back current row i */ 
    } 

    /* transform columns (each is running vertically with i) */ 
    for (j=0; j<8; j++) 
    { 
     for (i=0; i<8; i++) 
      in[i] = rows[i][j]; /* fill current column j */ 
     dct_1d(in, out, 8); /* transform current column j */ 
     for (i=0; i<8; i++) data[i][j] = out[i]; /* copy back current column j */ 
    } 
}