c - 这个二维 DCT 代码实际上是如何工作的？

Question

这是代码:

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];
    }
}

它取自listing2.c，位于 https://unix4lyfe.org/dct/

我只有 1 个问题，我们将行填充为 rows[j][i]，然后读出 rows[i][j]。根据 2D DCT 公式，我们转置 DCT 矩阵而不是实际数据。为什么要转置实际数据？

Answer 1

如果我假设 xpos 作为水平索引，ypos 作为垂直索引，则以下情况成立。

函数dct_1d(*,*,*);仅适用于一维数组(此处in和out )。您所迷惑的原因是这里的C中二维数组的索引杂技(特别是行) > 此处)。

通过在第一个 for block 中简单交换变量 i 和 j，可以将相同的代码重写如下，这将使得您正在尝试的物理感觉(请参阅评论):

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 */
    }
}