Question

我的计算机图形作业是仅使用绘制点的能力来实现OpenGL算法。

所以很明显我需要先让drawLine()工作才能画出其他东西。 drawLine()必须仅使用整数来完成。没有浮点。

这就是我的教导。基本上，线可以分为4个不同的类别，正陡，正浅，负陡和负浅。这是我应该画的图片：

expected result

这是我的程序绘制的图片：

actual result

颜色是为我们完成的。我们给出了顶点，我们需要使用Bresenham的Line算法根据起点和终点绘制线条。

这是我到目前为止所做的：

int dx = end.x - start.x;
int dy = end.y - start.y;

//initialize varibales
int d;
int dL;
int dU;

if (dy > 0){
        if (dy > dx){
                //+steep
                d = dy - 2*dx;
                dL = -2*dx;
                dU = 2*dy - 2*dx;

                for (int x = start.x, y = start.y; y <= end.y; y++){
                        Vertex v(x,y);
                        drawPoint(v);

                        if (d >= 1){
                                d += dL;
                        }else{
                                x++;
                                d += dU;
                        }
                }              
        } else {
                //+shallow
                d = 2*dy - dx;
                dL = 2*dy;
                dU = 2*dy - 2*dx;

                for (int x = start.x, y = start.y; x <= end.x; x++) {
                        Vertex v(x,y);
                        drawPoint(v);

                        // if choosing L, next y will stay the same, we only need
                        // to update d by dL
                        if (d <= 0) {
                                d += dL;
                        // otherwise choose U, y moves up 1
                        } else {
                                y++;
                                d += dU;
                        }
                }
        }
} else {
        if (-dy > dx){
                cout << "-steep\n";
                //-steep
                d = dy - 2*dx;
                //south
                dL = 2*dx;
                //southeast
                dU = 2*dy - 2*dx;

                for (int x = start.x, y = start.y; y >= end.y; --y){
                        Vertex v(x,y);
                        drawPoint(v);

                        //if choosing L, next x will stay the same, we only need
                        //to update d
                        if (d >= 1){
                                d -= dL;
                        } else {
                                x++;
                                d -= dU;
                        }
                }

        } else {
                cout << "-shallow\n";
                //-shallow
                d = 2*dy - dx;
                dL = 2*dy;
                dU = 2*dy - 2*dx;

                for (int x = start.x, y = start.y; x <= end.x; x++){
                        Vertex v(x,y);
                        drawPoint(v);

                        if (d >= 0){
                                d += dL;
                        } else {
                                --y;
                                d -= dU;
                        }
                }
        }
}

我知道我的错误会有些愚蠢，但我老实说无法弄清楚我做错了什么。为什么有些线条如上图所示不正确？

Answer 1

您可以在C ++中找到完整的代码，使用Bresenham算法在http://www.etechplanet.com/codesnippets/computer-graphics-draw-a-line-using-bresenham-algorithm.aspx处绘制一条线：

/*BRESENHAAM ALGORITHM FOR LINE DRAWING*/
#include<iostream.h>
#include<graphics.h>
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<math.h>
#include<dos.h>
void bhm_line(int,int,int,int,int);
void main()
{
 int ghdriver=DETECT,ghmode,errorcode,x1,x2,y1,y2;
 initgraph(&ghdriver,&ghmode,"..\\bgi");
 errorcode = graphresult();
 if(errorcode !=grOk)
 {
  cout<<"Graphics error:%s\n"<<grapherrormsg(errorcode);
  cout<<"Press any key to halt:";
  getch();
  exit(1);
 }
 clrscr();
 cout<<"Enter the coordinates (x1,y1): ";
 cin>>x1>>y1;
 cout<<"Enter the coordinates (x2,y2): ";
 cin>>x2>>y2;
 bhm_line(x1,y1,x2,y2,1);
 getch();
}
void bhm_line(int x1,int y1,int x2,int y2,int c)
{
 int x,y,dx,dy,dx1,dy1,px,py,xe,ye,i;
 dx=x2-x1;
 dy=y2-y1;
 dx1=fabs(dx);
 dy1=fabs(dy);
 px=2*dy1-dx1;
 py=2*dx1-dy1;
 if(dy1<=dx1)
 {
  if(dx>=0)
  {
   x=x1;
   y=y1;
   xe=x2;
  }
  else
  {
   x=x2;
   y=y2;
   xe=x1;
  }
  putpixel(x,y,c);
  for(i=0;x<xe;i++)
  {
   x=x+1;
   if(px<0)
   {
    px=px+2*dy1;
   }
   else
   {
    if((dx<0 && dy<0) || (dx>0 && dy>0))
    {
     y=y+1;
    }
    else
    {
     y=y-1;
    }
    px=px+2*(dy1-dx1);
   }
   delay(0);
   putpixel(x,y,c);
  }
 }
 else
 {
  if(dy>=0)
  {
   x=x1;
   y=y1;
   ye=y2;
  }
  else
  {
   x=x2;
   y=y2;
   ye=y1;
  }
  putpixel(x,y,c);
  for(i=0;y<ye;i++)
  {
   y=y+1;
   if(py<=0)
   {
    py=py+2*dx1;
   }
   else
   {
    if((dx<0 && dy<0) || (dx>0 && dy>0))
    {
     x=x+1;
    }
    else
    {
     x=x-1;
    }
    py=py+2*(dx1-dy1);
   }
   delay(0);
   putpixel(x,y,c);
  }
 }
}

Answer 2

如果有人想知道问题是什么，我仍然不知道它是什么。我最终做的是重新考虑我的代码，以便-shallow和-steep分别使用与+ shallow和+ steep相同的算法。在调整x，y坐标（否定x或y坐标）之后，当我去绘制它们时，我否定了我原来的否定，以便它绘制在正确的位置。

Answer 3

我用C ++实现了最初的Bresenham算法，并尝试了尽可能多的优化（尤其是从内部循环中删除IF的问题）。

它绘制的是线性缓冲区而不是表面，因此，此实现几乎与EFLA (Extremely Fast Line Algorithm)一样快（可能慢5％）。

#include <vector>
#include <math.h>
using namespace std;

vector<unsigned char> buffer;

int imageSide = 2048; // the width of the surface

struct Point2Di
{
    int x;
    int y;
    Point2Di(const int &x, const int &y): x(x), y(y){}
    Point2Di(){}

};

void drawLine(const Point2Di &p0, const Point2Di &p1)
{
    int dx = p1.x - p0.x;
    int dy = p1.y - p0.y;

    int dLong = abs(dx);
    int dShort = abs(dy);

    int offsetLong = dx > 0 ? 1 : -1;
    int offsetShort = dy > 0 ? imageSide : -imageSide;

    if(dLong < dShort)
    {
        swap(dShort, dLong);
        swap(offsetShort, offsetLong);
    }

    int error = dLong/2;
    int index = p0.y*imageSide + p0.x;
    const int offset[] = {offsetLong, offsetLong + offsetShort};
    const int abs_d[]  = {dShort, dShort - dLong};
    for(int i = 0; i <= dLong; ++i)
    {
        buffer[index] = 255;  // or a call to your painting method
        const int errorIsTooBig = error >= dLong;
        index += offset[errorIsTooBig];
        error += abs_d[errorIsTooBig];
    }
}

我正在使用的EFLA实现是：

void drawLine(Point2Di p0,  Point2Di p1)
{
    bool yLonger=false;
    int shortLen=p1.y-p0.y;
    int longLen=p1.x-p0.x;
    if (abs(shortLen)>abs(longLen)) {
        swap(shortLen, longLen);
        yLonger=true;
    }
    int decInc = longLen==0 ?  decInc=0 : ((shortLen << 16) / longLen);

    if (yLonger) {
        p0.y*=imageSide;
        p1.y*=imageSide;
        if (longLen>0)
            for (int j=0x8000+(p0.x<<16);p0.y<=p1.y;p0.y+=imageSide, j+=decInc)
                buffer[p0.y + (j >> 16)] = 255;  // or a call to your painting method
        else
            for (int j=0x8000+(p0.x<<16);p0.y>=p1.y;p0.y-=imageSide, j-=decInc)
                buffer[p0.y + (j >> 16)] = 255;  // or a call to your painting method
    }
    else
    {
        if (longLen>0)
            for (int j=0x8000+(p0.y<<16);p0.x<=p1.x;++p0.x, j+=decInc)
                buffer[(j >> 16) * imageSide + p0.x] = 255;  // or a call to your painting method
        else
            for (int j=0x8000+(p0.y<<16);p0.x>=p1.x;--p0.x, j-=decInc)
                buffer[(j >> 16) * imageSide + p0.x] = 255;  // or a call to your painting method
    }
}

用Bresenham线算法绘制线条

3 个答案: