Profiler 表示总时间的 50% 花费在这个函数中。你会如何优化它？ 它将 BMP 配色方案转换为 YUV。谢谢!

更新:平台为ARMV6(为iPhone编写)

#define Y_FROM_RGB(_r_,_g_,_b_) ( (  66 * _b_ + 129 * _g_ +  25 * _r_ + 128) >> 8) + 16
#define V_FROM_RGB(_r_,_g_,_b_) ( ( 112 * _b_ -  94 * _g_ -  18 * _r_ + 128) >> 10) + 128
#define U_FROM_RGB(_r_,_g_,_b_) ( ( -38 * _b_ -  74 * _g_ + 112 * _r_ + 128) >> 10) + 128

  /*!
 * \brief
 * Converts 24 bit image to YCrCb image channels
 * 
 * \param source
 * Source 24bit image pointer
 * 
 * \param source_width
 * Source image width
 * 
 * \param dest_Y
 * destination image Y component pointer
 * 
 * \param dest_scan_size_Y
 * destination image Y component line size
 * 
 * \param dest_U
 * destination image U component pointer
 * 
 * \param dest_scan_size_U
 * destination image U component line size
 * 
 * \param dest_V
 * destination image V component pointer
 * 
 * \param dest_scan_size_V
 * destination image V component line size
 * 
 * \param dest_width
 * Destination image width = source_width
 * 
 * \param dest_height
 * Destination image height = source image height
 *
 * Convert 24 bit image (source) with width (source_width)
 * to YCrCb image channels (dest_Y, dest_U, dest_V) with size (dest_width)x(dest_height), and line size
 * (dest_scan_size_Y, dest_scan_size_U, dest_scan_size_V) (in bytes)
 * 
 */
void ImageConvert_24_YUV420P(unsigned char * source, int source_width,
                            unsigned char * dest_Y, int dest_scan_size_Y,
                            unsigned char * dest_U, int dest_scan_size_U,
                            unsigned char * dest_V, int dest_scan_size_V,
                            int dest_width, int dest_height)
{
  int source_scan_size = source_width*3;

  int half_width = dest_width/2;

  //Y loop
  for (int y = 0; y < dest_height/2; y ++)
  {
    //Start of line
    unsigned char * source_scan = source;
    unsigned char * source_scan_next = source+source_scan_size;
    unsigned char * dest_scan_Y = dest_Y;
    unsigned char * dest_scan_U = dest_U;
    unsigned char * dest_scan_V = dest_V;

    //Do all pixels
    for (int x = 0; x < half_width; x++)
    {
      int R = source_scan[0];
      int G = source_scan[1];
      int B = source_scan[2];

      //Y
      int Y = Y_FROM_RGB(B, G, R);

      *dest_scan_Y = Y;
      source_scan += 3;
      dest_scan_Y += 1;

      int R1 = source_scan[0];
      int G1 = source_scan[1];
      int B1 = source_scan[2];

      //Y
      Y = Y_FROM_RGB(B1, G1, R1);

      R += (R1 + source_scan_next[0] + source_scan_next[3]);
      G += (G1 + source_scan_next[1] + source_scan_next[4]);
      B += (B1 + source_scan_next[2] + source_scan_next[5]);


      //YCrCb
      *dest_scan_Y = Y;
      *dest_scan_V = V_FROM_RGB(B, G, R);
      *dest_scan_U = U_FROM_RGB(B, G, R);

      source_scan += 3;
      dest_scan_Y += 1;
      dest_scan_U += 1;
      dest_scan_V += 1;
      source_scan_next += 6;
    };

    //scroll to next line
    source += source_scan_size;
    dest_Y += dest_scan_size_Y;
    dest_U += dest_scan_size_U;
    dest_V += dest_scan_size_V;

    //Start of line
    source_scan = source;
    dest_scan_Y = dest_Y;

    //Do all pixels
    for (int x = 0; x < half_width; x ++)
    {
      int R = source_scan[0];
      int G = source_scan[1];
      int B = source_scan[2];

      //Y
      int Y = Y_FROM_RGB(B, G, R);

      *dest_scan_Y = Y;
      source_scan += 3;
      dest_scan_Y += 1;

      R = source_scan[0];
      G = source_scan[1];
      B = source_scan[2];

      //Y
      Y = Y_FROM_RGB(B, G, R);
      *dest_scan_Y = Y;
      source_scan += 3;
      dest_scan_Y += 1;
    };

    source += source_scan_size;
    dest_Y += dest_scan_size_Y;
  };
};

最佳答案

除非我遗漏了什么，否则后续代码似乎在两个循环中都重复了，那么，为什么不通过这个循环一次呢？这可能需要对您的算法进行一些更改，但会提高性能。

for (int x = 0; x < half_width; x ++) 
{ 
  int R = source_scan[0]; 
  int G = source_scan[1]; 
  int B = source_scan[2]; 

  //Y 
  int Y = Y_FROM_RGB(B, G, R); 

  *dest_scan_Y = Y; 
  source_scan += 3; 
  dest_scan_Y += 1; 

  R = source_scan[0]; 
  G = source_scan[1]; 
  B = source_scan[2]; 

但是，在做任何事情之前，将两个内部循环移动到单独的函数中，然后运行您的探查器，看看您是否在一个函数上花费的时间比另一个函数多。

你在这个函数中有三个循环，你不知道哪个部分是你实际花时间的地方。所以在进行任何优化之前确定这一点，否则您可能会发现您正在修复错误的部分。

关于c++ - 如何优化这段代码？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3439594/