C++ 三角形光栅化

Question

我正在尝试修复此三角形光栅化器，但无法使其正常工作。由于某种原因，它只绘制了一半的三角形。

void DrawTriangle(Point2D p0, Point2D p1, Point2D p2)
{
    Point2D Top, Middle, Bottom;
    bool MiddleIsLeft;

    if (p0.y < p1.y)                    // case: 1, 2, 5
    {
        if (p0.y < p2.y)                // case: 1, 2
        {
            if (p1.y < p2.y)            // case: 1
            {
                Top = p0;
                Middle = p1;
                Bottom = p2;
                MiddleIsLeft = true;
            }
            else                        // case: 2
            {
                Top = p0;
                Middle = p2;
                Bottom = p1;
                MiddleIsLeft = false;
            }
        }
        else                            // case: 5
        {
            Top = p2;
            Middle = p0;
            Bottom = p1;
            MiddleIsLeft = true;                
        }
    }
    else                        // case: 3, 4, 6
    {
        if (p0.y < p2.y)        // case: 4
        {
            Top = p1;
            Middle = p0;
            Bottom = p2;
            MiddleIsLeft = false;
        }
        else                    // case: 3, 6
        {
            if (p1.y < p2.y)    // case: 3
            {
                Top = p1;
                Middle = p2;
                Bottom = p0;
                MiddleIsLeft = true;
            }
            else                // case 6
            {
                Top = p2;
                Middle = p1;
                Bottom = p0;
                MiddleIsLeft = false;
            }
        }
    }

    float xLeft, xRight;
    xLeft = xRight = Top.x;
    float mLeft, mRight;
    // Region 1
    if(MiddleIsLeft)
    {
        mLeft = (Top.x - Middle.x) / (Top.y - Middle.y);
        mRight = (Top.x - Bottom.x) / (Top.y - Bottom.y);
    }
    else
    {
        mLeft = (Top.x - Bottom.x) / (Top.y - Bottom.y);
        mRight = (Middle.x - Top.x) / (Middle.y - Top.y);
    }
    int finalY;
    float Tleft, Tright;
    for (int y = ceil(Top.y); y < (int)Middle.y; y++)
    {        
        Tleft=float(Top.y-y)/(Top.y-Middle.y);
        Tright=float(Top.y-y)/(Top.y-Bottom.y);
        for (int x = ceil(xLeft); x <= ceil(xRight) - 1 ; x++)
        {
            FrameBuffer::SetPixel(x, y, p0.r,p0.g,p0.b);

        }  
        xLeft += mLeft;
        xRight += mRight;
        finalY = y;
    }

    // Region 2 
    if (MiddleIsLeft)
    {
        mLeft = (Bottom.x - Middle.x) / (Bottom.y - Middle.y);
    }
    else
    {
        mRight = (Middle.x - Bottom.x) / (Middle.y - Bottom.y);
    }

    for (int y = Middle.y; y <= ceil(Bottom.y) - 1; y++)
    {
        Tleft=float(Bottom.y-y)/(Bottom.y-Middle.y);
        Tright=float(Top.y-y)/(Top.y-Bottom.y);
        for (int x = ceil(xLeft); x <= ceil(xRight) - 1; x++)
        {
            FrameBuffer::SetPixel(x, y, p0.r,p0.g,p0.b);
        }
        xLeft += mLeft;
        xRight += mRight; 

    }
}

这是我用它来绘制形状时发生的情况。

当我禁用第二个区域时，所有那些奇怪的三角形都会消失。

线框模式完美运行，因此消除了除三角形光栅器之外的所有其他可能性。

Answer 1

我有点迷失在你的实现中，但这就是我所做的（我有一个稍微复杂一点的版本，用于任意凸多边形，而不仅仅是三角形），我认为除了 Bresenham's algorithm 它非常简单（实际上是算法也很简单）：

#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;
}

输出：

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

Answer 2

原始代码仅适用于具有逆时针缠绕的三角形，因为顶部的 if-else 语句确定中间是左还是右。可能是未绘制的三角形的绕组错误。

此堆栈溢出显示如何Determine winding of a 2D triangles after triangulation

原始代码速度很快，因为它不会将线的点保存在临时内存缓冲区中。即使考虑到这一点，似乎也有点过于复杂，但这是另一个问题。

Answer 3

以下代码在您的实现中：

if (p0.y < p1.y)                    // case: 1, 2, 5

    {
        if (p0.y < p2.y)                // case: 1, 2
        {
            if (p1.y < p2.y)            // case: 1
            {
                Top = p0;
                Middle = p1;
                Bottom = p2;
                MiddleIsLeft = true;
            }
            else                        // case: 2
            {
                Top = p0;
                Middle = p2;
                Bottom = p1;
                MiddleIsLeft = false;
            }
        }

这个 else 语句意味着 p2.y（或 Middle）可以等于 p1.y（或 Bottom）。如果这是真的，那么当区域 2 运行时

if (MiddleIsLeft)

    {
        mLeft = (Bottom.x - Middle.x) / (Bottom.y - Middle.y);
    }
    else
    {
        mRight = (Middle.x - Bottom.x) / (Middle.y - Bottom.y);
    }

else 行将被零除，这是不可能的。