解释 ARM Neon 图像采样

Question

我正在尝试编写 OpenCV 的 cv::resize() 的更好版本，但我遇到了一个交叉代码：https://github.com/rmaz/NEON-Image-Downscaling/blob/master/ImageResize/BDPViewController.m 该代码用于将图像下采样 2，但我无法获得算法。我想首先将该算法转换为 C，然后尝试对其进行修改以用于学习目的。是否也很容易将其转换为任意大小的下采样？

函数是：

static void inline resizeRow(uint32_t *dst, uint32_t *src, uint32_t pixelsPerRow)
{
    const uint32_t * rowB = src + pixelsPerRow;

    // force the number of pixels per row to a multiple of 8
    pixelsPerRow = 8 * (pixelsPerRow / 8);

    __asm__ volatile("Lresizeloop: \n" // start loop
                     "vld1.32 {d0-d3}, [%1]! \n" // load 8 pixels from the top row
                     "vld1.32 {d4-d7}, [%2]! \n" // load 8 pixels from the bottom row
                     "vhadd.u8 q0, q0, q2 \n" // average the pixels vertically
                     "vhadd.u8 q1, q1, q3 \n"
                     "vtrn.32 q0, q2 \n" // transpose to put the horizontally adjacent pixels in different registers
                     "vtrn.32 q1, q3 \n"
                     "vhadd.u8 q0, q0, q2 \n" // average the pixels horizontally
                     "vhadd.u8 q1, q1, q3 \n"
                     "vtrn.32 d0, d1 \n" // fill the registers with pixels
                     "vtrn.32 d2, d3 \n"
                     "vswp d1, d2 \n"
                     "vst1.64 {d0-d1}, [%0]! \n" // store the result
                     "subs %3, %3, #8 \n" // subtract 8 from the pixel count
                     "bne Lresizeloop \n" // repeat until the row is complete
: "=r"(dst), "=r"(src), "=r"(rowB), "=r"(pixelsPerRow)
: "0"(dst), "1"(src), "2"(rowB), "3"(pixelsPerRow)
: "q0", "q1", "q2", "q3", "cc"
);
}

To call it:

 // downscale the image in place
    for (size_t rowIndex = 0; rowIndex < height; rowIndex+=2)
    {
        void *sourceRow = (uint8_t *)buffer + rowIndex * bytesPerRow;
        void *destRow = (uint8_t *)buffer + (rowIndex / 2) * bytesPerRow;
        resizeRow(destRow, sourceRow, width);
    }

Answer 1

算法非常简单。它从当前行读取 8 个像素，从下一行读取 8 个像素。然后它使用 vhadd（减半相加）指令垂直平均 8 个像素。然后它转置像素的位置，以便水平相邻的像素对现在位于单独的寄存器中（垂直排列）。然后它执行另一组减半加法以将它们平均在一起。然后将结果再次转换以将它们放在原始位置并写入目的地。可以重写该算法以处理不同的缩放积分大小，但正如所写的那样，它只能通过平均进行 2x2 到 1 的缩减。这是等效的 C 代码：

static void inline resizeRow(uint32_t *dst, uint32_t *src, uint32_t pixelsPerRow)
{
    uint8_t * pSrc8 = (uint8_t *)src;
    uint8_t * pDest8 = (uint8_t *)dst;
    int stride = pixelsPerRow * sizeof(uint32_t);
    int x;
    int r, g, b, a;

    for (x=0; x<pixelsPerRow; x++)
    {
       r = pSrc8[0] + pSrc8[4] + pSrc8[stride+0] + pSrc8[stride+4];
       g = pSrc8[1] + pSrc8[5] + pSrc8[stride+1] + pSrc8[stride+5];
       b = pSrc8[2] + pSrc8[6] + pSrc8[stride+2] + pSrc8[stride+6];
       a = pSrc8[3] + pSrc8[7] + pSrc8[stride+3] + pSrc8[stride+7];
       pDest8[0] = (uint8_t)((r + 2)/4); // average with rounding
       pDest8[1] = (uint8_t)((g + 2)/4);
       pDest8[2] = (uint8_t)((b + 2)/4);
       pDest8[3] = (uint8_t)((a + 2)/4);
       pSrc8 += 8; // skip forward 2 source pixels
       pDest8 += 4; // skip forward 1 destination pixel
    }