C++ triangle rasterization

可紊 提交于 2019-11-28 07:50:42
Alexey Frunze

I kind of got lost in your implementation, but here's what I do (I have a slightly more complex version for arbitrary convex polygons, not just triangles) and I think apart from the Bresenham's algorithm it's very simple (actually the algorithm is simple too):

#include <stddef.h>
#include <limits.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>

#define SCREEN_HEIGHT 22
#define SCREEN_WIDTH  78

// Simulated frame buffer
char Screen[SCREEN_HEIGHT][SCREEN_WIDTH];

void SetPixel(long x, long y, char color)
{
  if ((x < 0) || (x >= SCREEN_WIDTH) ||
      (y < 0) || (y >= SCREEN_HEIGHT))
  {
    return;
  }

  Screen[y][x] = color;
}

void Visualize(void)
{
  long x, y;

  for (y = 0; y < SCREEN_HEIGHT; y++)
  {
    for (x = 0; x < SCREEN_WIDTH; x++)
    {
      printf("%c", Screen[y][x]);
    }

    printf("\n");
  }
}

typedef struct
{
  long x, y;
  unsigned char color;
} Point2D;


// min X and max X for every horizontal line within the triangle
long ContourX[SCREEN_HEIGHT][2];

#define ABS(x) ((x >= 0) ? x : -x)

// Scans a side of a triangle setting min X and max X in ContourX[][]
// (using the Bresenham's line drawing algorithm).
void ScanLine(long x1, long y1, long x2, long y2)
{
  long sx, sy, dx1, dy1, dx2, dy2, x, y, m, n, k, cnt;

  sx = x2 - x1;
  sy = y2 - y1;

  if (sx > 0) dx1 = 1;
  else if (sx < 0) dx1 = -1;
  else dx1 = 0;

  if (sy > 0) dy1 = 1;
  else if (sy < 0) dy1 = -1;
  else dy1 = 0;

  m = ABS(sx);
  n = ABS(sy);
  dx2 = dx1;
  dy2 = 0;

  if (m < n)
  {
    m = ABS(sy);
    n = ABS(sx);
    dx2 = 0;
    dy2 = dy1;
  }

  x = x1; y = y1;
  cnt = m + 1;
  k = n / 2;

  while (cnt--)
  {
    if ((y >= 0) && (y < SCREEN_HEIGHT))
    {
      if (x < ContourX[y][0]) ContourX[y][0] = x;
      if (x > ContourX[y][1]) ContourX[y][1] = x;
    }

    k += n;
    if (k < m)
    {
      x += dx2;
      y += dy2;
    }
    else
    {
      k -= m;
      x += dx1;
      y += dy1;
    }
  }
}

void DrawTriangle(Point2D p0, Point2D p1, Point2D p2)
{
  int y;

  for (y = 0; y < SCREEN_HEIGHT; y++)
  {
    ContourX[y][0] = LONG_MAX; // min X
    ContourX[y][1] = LONG_MIN; // max X
  }

  ScanLine(p0.x, p0.y, p1.x, p1.y);
  ScanLine(p1.x, p1.y, p2.x, p2.y);
  ScanLine(p2.x, p2.y, p0.x, p0.y);

  for (y = 0; y < SCREEN_HEIGHT; y++)
  {
    if (ContourX[y][1] >= ContourX[y][0])
    {
      long x = ContourX[y][0];
      long len = 1 + ContourX[y][1] - ContourX[y][0];

      // Can draw a horizontal line instead of individual pixels here
      while (len--)
      {
        SetPixel(x++, y, p0.color);
      }
    }
  }
}

int main(void)
{
  Point2D p0, p1, p2;

  // clear the screen
  memset(Screen, ' ', sizeof(Screen));

  // generate random triangle coordinates
  srand((unsigned)time(NULL));

  p0.x = rand() % SCREEN_WIDTH;
  p0.y = rand() % SCREEN_HEIGHT;

  p1.x = rand() % SCREEN_WIDTH;
  p1.y = rand() % SCREEN_HEIGHT;

  p2.x = rand() % SCREEN_WIDTH;
  p2.y = rand() % SCREEN_HEIGHT;

  // draw the triangle
  p0.color = '1';
  DrawTriangle(p0, p1, p2);

  // also draw the triangle's vertices
  SetPixel(p0.x, p0.y, '*');
  SetPixel(p1.x, p1.y, '*');
  SetPixel(p2.x, p2.y, '*');

  Visualize();

  return 0;
}

Output:

   *111111
    1111111111111
      111111111111111111
         1111111111111111111111
           111111111111111111111111111
             11111111111111111111111111111111
                111111111111111111111111111111111111
                  11111111111111111111111111111111111111111
                    111111111111111111111111111111111111111*
                       11111111111111111111111111111111111
                         1111111111111111111111111111111
                            111111111111111111111111111
                              11111111111111111111111
                                1111111111111111111
                                   11111111111111
                                     11111111111
                                       1111111
                                          1*
Treb

The original code will only work properly with triangles that have counter-clockwise winding because of the if-else statements on top that determines whether middle is left or right. It could be that the triangles which aren't drawing have the wrong winding.

This stack overflow shows how to Determine winding of a 2D triangles after triangulation

The original code is fast because it doesn't save the points of the line in a temporary memory buffer. Seems a bit over-complicated even given that, but that's another problem.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!