【问题标题】:HOWTO draw circles, arcs and vector graphics in SDL?如何在 SDL 中绘制圆、弧和矢量图形?
【发布时间】:2016-11-15 00:03:41
【问题描述】:

(我正在使用 SDL2)

SDL is a relatively small library for "low level access to audio, keyboard, mouse, joystick, and graphics hardware via OpenGL and Direct3D" 它用于游戏开发,在我的情况下,用作简单的视听输出和鼠标+键盘输入。它不是像 GTK、Qt、wxWindows 等那样的“工具包”,但它是跨平台的。

But the only way I can find to draw a shape is with the line, rect and pixel functions.

除了使用三角函数或“圆方程”之外,我如何绘制曲线?一般的矢量图呢?

SDL 是一个合适的起点吗?还是我应该看看其他地方?

【问题讨论】:

    标签: c sdl sdl-2 vector-graphics


    【解决方案1】:

    这是上面提到的中点圆算法的一个例子。它不需要数学库,而且速度非常快。 (在大约 500 微秒内渲染)这是 Windows 使用/用于光栅化圆圈的方法。

    void DrawCircle(SDL_Renderer * renderer, int32_t centreX, int32_t centreY, int32_t radius)
    {
       const int32_t diameter = (radius * 2);
    
       int32_t x = (radius - 1);
       int32_t y = 0;
       int32_t tx = 1;
       int32_t ty = 1;
       int32_t error = (tx - diameter);
    
       while (x >= y)
       {
          //  Each of the following renders an octant of the circle
          SDL_RenderDrawPoint(renderer, centreX + x, centreY - y);
          SDL_RenderDrawPoint(renderer, centreX + x, centreY + y);
          SDL_RenderDrawPoint(renderer, centreX - x, centreY - y);
          SDL_RenderDrawPoint(renderer, centreX - x, centreY + y);
          SDL_RenderDrawPoint(renderer, centreX + y, centreY - x);
          SDL_RenderDrawPoint(renderer, centreX + y, centreY + x);
          SDL_RenderDrawPoint(renderer, centreX - y, centreY - x);
          SDL_RenderDrawPoint(renderer, centreX - y, centreY + x);
    
          if (error <= 0)
          {
             ++y;
             error += ty;
             ty += 2;
          }
    
          if (error > 0)
          {
             --x;
             tx += 2;
             error += (tx - diameter);
          }
       }
    }
    

    【讨论】:

      【解决方案2】:

      如果你想写自己的画圆函数,那么我建议通过绘制像素将midpoint algorithm适配到SDL2。

      曲线可以类似地完成,但会使用更多的椭圆绘制算法。

      实际的矢量图形开始变得更加复杂,您可能必须找到可以渲染 SVG 文件的东西,我不确定 SDL2 是否有很多选项。

      但是,如果您只想拥有可以使用的功能,我建议您直接转至 SDL2_gfx。它已经实现了更多功能供您使用。

      【讨论】:

        【解决方案3】:

        SDL 允许第三方库在纹理上绘制。如果 cairo 是可取的,它可以用在这样的函数中:

        cairo_t*cb(cairo_t*cr)
        {cairo_set_source_rgb(cr, 1.0, 1.0, 1.0);
         cairo_rectangle(cr, 10, 20, 128, 128);
         cairo_stroke(cr);
         return cr;
        }
        

        那么cb可以传递给这个函数:

        cairo_t*cai(SDL_Window*w,SDL_Renderer*r,cairo_t*(*f)(cairo_t*))
        {int width, height, pitch;void *pixels;
         SDL_GetWindowSize(w, &width, &height);
         SDL_Texture*t=SDL_CreateTexture(r,SDL_PIXELFORMAT_ARGB8888,SDL_TEXTUREACCESS_STREAMING,width,height);
         SDL_LockTexture(t, NULL, &pixels, &pitch);
         cairo_surface_t *cs=cairo_image_surface_create_for_data(pixels,CAIRO_FORMAT_ARGB32,width,height,pitch);
         cairo_t*s=cairo_create(cs);
         cairo_t*fr=f(s);SDL_UnlockTexture(t);SDL_RenderCopy(r,t,NULL,NULL);SDL_RenderPresent(r);
         return fr;
        }
        

        【讨论】:

          【解决方案4】:

          如果您想在没有第三方库的情况下制作圆形或椭圆形,请包含 math.h 并使用我编写的下面的函数。它会很好地绘制别名椭圆或圆形。在 SDL 2.0.2 上测试并且可以工作。它绘制一个象限弧,并镜像其他弧,减少对 cosf 和 sinf 的调用。

          //draw one quadrant arc, and mirror the other 4 quadrants
          void sdl_ellipse(SDL_Renderer* r, int x0, int y0, int radiusX, int radiusY)
          {
              float pi  = 3.14159265358979323846264338327950288419716939937510;
              float pih = pi / 2.0; //half of pi
          
              //drew  28 lines with   4x4  circle with precision of 150 0ms
              //drew 132 lines with  25x14 circle with precision of 150 0ms
              //drew 152 lines with 100x50 circle with precision of 150 3ms
              const int prec = 27; // precision value; value of 1 will draw a diamond, 27 makes pretty smooth circles.
              float theta = 0;     // angle that will be increased each loop
          
              //starting point
              int x  = (float)radiusX * cos(theta);//start point
              int y  = (float)radiusY * sin(theta);//start point
              int x1 = x;
              int y1 = y;
          
              //repeat until theta >= 90;
              float step = pih/(float)prec; // amount to add to theta each time (degrees)
              for(theta=step;  theta <= pih;  theta+=step)//step through only a 90 arc (1 quadrant)
              {
                  //get new point location
                  x1 = (float)radiusX * cosf(theta) + 0.5; //new point (+.5 is a quick rounding method)
                  y1 = (float)radiusY * sinf(theta) + 0.5; //new point (+.5 is a quick rounding method)
          
                  //draw line from previous point to new point, ONLY if point incremented
                  if( (x != x1) || (y != y1) )//only draw if coordinate changed
                  {
                      SDL_RenderDrawLine(r, x0 + x, y0 - y,    x0 + x1, y0 - y1 );//quadrant TR
                      SDL_RenderDrawLine(r, x0 - x, y0 - y,    x0 - x1, y0 - y1 );//quadrant TL
                      SDL_RenderDrawLine(r, x0 - x, y0 + y,    x0 - x1, y0 + y1 );//quadrant BL
                      SDL_RenderDrawLine(r, x0 + x, y0 + y,    x0 + x1, y0 + y1 );//quadrant BR
                  }
                  //save previous points
                  x = x1;//save new previous point
                  y = y1;//save new previous point
              }
              //arc did not finish because of rounding, so finish the arc
              if(x!=0)
              {
                  x=0;
                  SDL_RenderDrawLine(r, x0 + x, y0 - y,    x0 + x1, y0 - y1 );//quadrant TR
                  SDL_RenderDrawLine(r, x0 - x, y0 - y,    x0 - x1, y0 - y1 );//quadrant TL
                  SDL_RenderDrawLine(r, x0 - x, y0 + y,    x0 - x1, y0 + y1 );//quadrant BL
                  SDL_RenderDrawLine(r, x0 + x, y0 + y,    x0 + x1, y0 + y1 );//quadrant BR
              }
          }
          

          【讨论】:

          • 如果prec 基于半径值而不是始终相同会更好吗?这将提高大椭圆的质量,但保持较小椭圆的速度。
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