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can the CImg library draw thick lines

开发者 https://www.devze.com 2023-02-24 19:32 出处:网络
I have been using the CImg library, and 开发者_StackOverflow社区have been pleased with how easy it is to integrate and use. However, I now want to draw thick lines (i.e., more than one pixel thick). I

I have been using the CImg library, and 开发者_StackOverflow社区have been pleased with how easy it is to integrate and use. However, I now want to draw thick lines (i.e., more than one pixel thick). It is not clear from the API documentation of the draw_line function (here) how this can be done. A second version of the function (just below the first in the documentation) even takes a texture as input, but again no width. It seems strange that such a comprehensive library would not have this feature. Perhaps it's supposed to be done using some kind of transformation? I know I could do it using a polygon (i.e., a rectangle where I would compute the corners of the polygon using a normal to the line), but I fear that would be significantly slower.


Apparently, it is not possible 'out-of-the-box', but creating your own routine that calls multiple times the 'draw_line()' routine of CImg, with one or two pixels shifts should give you the result you want, without much work.


This function can be used to draw thick lines as polygons.

void draw_line(cimg_library::CImg<uint8_t>& image,
    const int x1, const int y1,
    const int x2, const int y2,
    const uint8_t* const color,
    const unsigned int line_width)
{
    if (x1 == x2 && y1 == y2) {
        return;
    }
    // Convert line (p1, p2) to polygon (pa, pb, pc, pd)
    const double x_diff = std::abs(x1 - x2);
    const double y_diff = std::abs(y1 - y2);
    const double w_diff = line_width / 2.0;

    // Triangle between pa and p1: x_adj^2 + y_adj^2 = w_diff^2
    // Triangle between p1 and p2: x_diff^2 + y_diff^2 = length^2 
    // Similar triangles: y_adj / x_diff = x_adj / y_diff = w_diff / length
    // -> y_adj / x_diff = w_diff / sqrt(x_diff^2 + y_diff^2) 
    const int x_adj = y_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));
    const int y_adj = x_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));

    // Points are listed in clockwise order, starting from top-left
    cimg_library::CImg<int> points(4, 2);
    points(0, 0) = x1 - x_adj;
    points(0, 1) = y1 + y_adj;
    points(1, 0) = x1 + x_adj;
    points(1, 1) = y1 - y_adj;
    points(2, 0) = x2 + x_adj;
    points(2, 1) = y2 - y_adj;
    points(3, 0) = x2 - x_adj;
    points(3, 1) = y2 + y_adj;

    image.draw_polygon(points, color);
}

Benchmarks with line_width 20 and 3 colors. First time is using this function, second time is drawing single 1 px wide line using image.draw_line().

  • 1000,1000 to 2000,2000: 216 µs / 123 µs
  • 2000,2000 to 8000,4000: 588 µs / 151 µs
  • 3000,1000 to 3020,1000: 21 µs / 5 µs


Basically this code does the same as @vll's answer, but also handles the case when (x1-x2)/(y1-y2) < 0 (I remove the abs function).

void draw_line(cimg_library::CImg<uint8_t>& image,
      const int x1, const int y1,
      const int x2, const int y2,
      const uint8_t* const color,
      const uint8_t line_width,
      const double opacity=1.0)
   {
      if (x1 == x2 && y1 == y2) {
         return;
      }
      // Convert line (p1, p2) to polygon (pa, pb, pc, pd)
      const double x_diff = (x1 - x2);
      const double y_diff = (y1 - y2);
      const double w_diff = line_width / 2.0;

      // Triangle between pa and p1: x_adj^2 + y_adj^2 = w_diff^2
      // Triangle between p1 and p2: x_diff^2 + y_diff^2 = length^2 
      // Similar triangles: y_adj / x_diff = x_adj / y_diff = w_diff / length
      // -> y_adj / x_diff = w_diff / sqrt(x_diff^2 + y_diff^2) 
      const int x_adj = y_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));
      const int y_adj = x_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));

      // Points are listed in clockwise order, starting from top-left
      cimg_library::CImg<int> points(4, 2);
      points(0, 0) = x1 - x_adj;
      points(0, 1) = y1 + y_adj;
      points(1, 0) = x1 + x_adj;
      points(1, 1) = y1 - y_adj;
      points(2, 0) = x2 + x_adj;
      points(2, 1) = y2 - y_adj;
      points(3, 0) = x2 - x_adj;
      points(3, 1) = y2 + y_adj;

      image.draw_polygon(points, color, opacity);
   }
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