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;
}
}
当我用它绘制形状时会发生什么。
当我禁用第二个区域时,所有那些奇怪的三角形都会消失。
线框模式工作正常,因此除了三角形光栅化器之外,它还消除了所有其他可能性。
答案 0 :(得分:15)
我有点迷失在你的实现中,但这就是我所做的(我有一个稍微复杂的版本,任意凸多边形,而不仅仅是三角形)我认为除了Bresenham's algorithm它很简单(实际上)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*
答案 1 :(得分:3)
原始代码仅适用于逆时针缠绕的三角形,因为顶部的if-else语句决定了中间是左还是右。可能是未绘制的三角形有错误的缠绕。
此堆栈溢出显示如何Determine winding of a 2D triangles after triangulation
原始代码很快,因为它不会将行的点保存在临时内存缓冲区中。即使考虑到这一点,似乎有点过于复杂,但这是另一个问题。