计算机gaphics c ++ [英] computer gaphics c++

问题描述

用于裁剪的c ++代码
窗户夹线

c++ code for line clipping
a window clips lines

#include<stdio.h>
#include<conio.h>
#include<dos.h>
#include<graphics.h>
#include<stdlib.h>
typedef struct coord
{
int x,y;
char code[4];
}PT;
void drawwindow();
void drawline(PT p1,PT p2,int c1);
PT setcode(PT p);
int visibility(PT p1,PT p2);
PT resetendpt(PT p1,PT p2);
void main()
{
int gd=DETECT,gm,v;
PT p1,p2,ptemp;
initgraph(&gd,&gm,"");
cleardevice();
printf("\t\tCOHEN-SUTHERLAND LINE CLIPPING ALGORITHM");
printf("\nEnter the two end points p1(x,y)");
scanf("%d%d",&p1.x,&p1.y);
printf("Enter the two end points p2(x,y):");
scanf("%d%d",&p2.x,&p2.y);
cleardevice();
printf("\t\tCLIPPING WINDOW");
drawwindow();
getch();
printf("\tCOHEN-SUTHERLAND LINE CLIPPING ALGORITHM");
printf("\n\t\tBEFORE CLIPPING");
drawline(p1,p2,4);
getch();
p1=setcode(p1);
p2=setcode(p2);
v=visibility(p1,p2);
switch(v)
{
  case 0:
   cleardevice();
   //printf("\t\tCLIPPING WINDOW");
   drawwindow();
   drawline(p1,p2,15);
  break;
 case 1:
   cleardevice();
   drawwindow();
  break;
  case 2:
   cleardevice();
   p1=resetendpt(p1,p2);
   p2=resetendpt(p2,p1);
   drawwindow();
   printf("\tCOHEN-SUTHERLAND LINE CLIPPING ALGORITHM");
   printf("\n\t\tAFTER CLIPPING");
   drawline(p1,p2,15);
  break;
 }
 getch();
 closegraph();
//return(0);
}
void drawwindow()
{
 setcolor(RED);
 setlinestyle(DOTTED_LINE, 1, 1);
 line(150,100,100,100);
 line(150,100,150,50);
 line(450,100,500,100);
 line(450,100,450,50);
 line(450,350,500,350);
 line(450,350,450,400);
 line(150,350,150,400);
 line(150,350,100,350) ;
 line(150,100,100,100);
 line(150,100,150,50);
 setlinestyle(SOLID_LINE, 1, 1);
 line(150,100,450,100);
 line(450,100,450,350);
 line(450,350,150,350);
 line(150,350,150,100);
 
 outtextxy(450,30,"1001");
 outtextxy(470,200,"1000");
 outtextxy(470,370,"1010");
 outtextxy(300,370,"0010");
 outtextxy(120,370,"0110");
 outtextxy(120,200,"0100");
 outtextxy(120,30,"0101");
 outtextxy(300,30,"0001");
 outtextxy(300,200,"0000");
}
void drawline(PT p1,PT p2,int c1)
{
 setcolor(c1);
 line(p1.x,p1.y,p2.x,p2.y);
}
 
PT setcode(PT p)
{
 PT ptemp;
 if(p.y<100)
  ptemp.code[0]='1';
 else
  ptemp.code[0]='0';
 if(p.y>350)
  ptemp.code[1]='1';
 else
  ptemp.code[1]='0';
 if(p.x>200)
  ptemp.code[2]='1';
 else
  ptemp.code[2]='0';
 if(p.x<150)
  ptemp.code[3]='1';
 else
  ptemp.code[3]='0';
  ptemp.x=p.x;
  ptemp.y=p.y;
 return(ptemp);
}
int visibility(PT p1,PT p2)
{
 int i,flag=0;
 for(i=0;i<4;i++)
 {
  if((p1.code[i]!='0')||(p2.code[i]!='0'))
  flag=1;
 }
 if(flag==0)
 return(0);
 for(i=0;i<4;i++)
 {
  if((p1.code[i]==p2.code[i])&&(p1.code[i]=='1'))
  flag=0;
 }
 if(flag==0)
 return(1);
 return(2);
}
PT resetendpt(PT p1,PT p2)
{
 PT temp;
 int x,y,i;
 float m,k;
 if(p1.code[3]=='1')
 x=150;
 if(p1.code[2]=='1')
 x=450;
 if((p1.code[3]=='1')||(p1.code[2]=='1'))
 {
  m=(float)(p2.y-p1.y)/(p2.x-p1.x);
  k=(p1.y+(m*(x-p1.x)));
  temp.y=k;
  temp.x=x;
  for(i=0;i<4;i++)
  {
   temp.code[i]=p1.code[i];
   if(temp.y<=350 && temp.y>=100)
    return(temp);
  }
  if(p1.code[0]=='1')
  y=100;
  if(p1.code[1]=='1')
  y=350;
  if((p1.code[0]=='1')||(p1.code[1]=='1'))
  {
   m=(float)(p2.y-p1.y)/(p2.x-p1.x);
   k=(float)p1.x+(float)(y-p1.y)/m;
   temp.y=y;
   temp.x=k;
   for(i=0;i<4;i++)
   temp.code[i]=p1.code[i];
   return(temp);
  }
 }
 else
 return(p1);
 return(p2);
}

我尝试了此操作，但没有成功

i tried this but it didnt work

推荐答案

我不知道您遇到了哪个特定问题，因为它们有几处需要改进. br/>
1)在您的setcode函数中，有一项针对p.x≥200的测试，显然应该读取为> =450.这可能是您遇到的麻烦.

2)在字符数组中存储属性的左，右，上，下不是一个好主意，因为它使您无法使用Cohen-Sutherland算法的巨大优势:能够通过按位OR-排除两个重要的情况运算和与运算.相反，您应该将它们作为位标志存储在int变量中.例如:

I don''t know which particular problem you are stumbling over, as their are a couple of things that need improvement.

1) In your setcode function there is one test for p.x >= 200, which should obviously read >= 450. That is probably the trouble you have.

2) Storing the attributes left, right, top, bottom in a character array is no good idea as it deprives you from the big advantage of the Cohen-Sutherland algorithm: Being able to rule out two important cases with a bit-wise OR-operation and AND-operation. Instead you should store them as bit-flags in an int variable. For example:

const int RIGHT  = 8;
const int LEFT   = 4;
const int BOTTOM = 2;
const int TOP    = 1;

int code = 0;
if (x < 150)
    code |= LEFT;
if (x > 450)
    code |= RIGHT;
....

现在，您可以通过询问:
检查两个点是否都在剪切矩形的同一侧.

Now you can check if both points are on the same side of the clipping rectangle by asking:

int code1 = getcode (pt1);
int code2 = getcode (pt2);
if ((code1 & code2) != 0)
    // both points are on one side and hence the
    // clipped line is empty
    return false;
if ((code1 | code2) == 0)
    // both points are inside the clipping rectange
    // so we can return the entire line
    ....

3)当然，您想以更通用的方式编写代码，
使用变量作为裁剪矩形的坐标.

4)您的resetendpt函数相对于原始的Cohen-Sutherland算法有所简化，并且存在以下问题:当两个点的x或y坐标相等时，将导致被零除.在原始算法中，这些情况已被我在(2)中显示的测试排除.

我建议您在Wikipedia或其他来源中查找原始算法，并研究它们的完成方式.当然，我可以在此处粘贴一段可以解决您的工作的代码，但是我认为作为一项家庭作业，重要的是您必须自己完成这项工作.如果遇到问题，请使用改善问题按钮告诉我们该问题，您肯定会在此过程中得到进一步的提示.

3) Of course you want to write your code in a more general way and
use variables for the coordinates of your clipping rectangle.

4) Your resetendpt function is somewhat simplified from the original Cohen-Sutherland algorithm and has the problem that when the x- or y-coordinates of both points are equal that will lead to division by zero. In the original algorithm, these cases have been excluded by the tests that I showed in (2).

I recommend you to lookup the original algorithm in Wikipedia or some other source and study, how it is done their. Of course, I could paste here a piece of code that solves your work, but I think as a homework assignment it is important that you do that by yourself. If you run into problems, please use the improve question button to tell us the problem, and you sure will get a further hint along the way.

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