Want to win a PS4? Go Premium and enter to win our High-Tech Treats giveaway. Enter to Win

x
?
Solved

Convert C code to C#

Posted on 2009-03-29
6
Medium Priority
?
536 Views
Last Modified: 2012-08-13
I need to convert this C code to C#

My conversion identifies my polygon as ccw. I know it is cw.
Did I make a mistake converting from C to C#?
I noted the assumption that the last point is not repeated and have tried my code with n-1, but still returns ccw.
Also, my polygon is in the Western Hemisphere, meaning that Longitude is expressed as a negative number.
I have tried all the solutions to deal with the negative number I can think of.
Thanks for any help,
Michael


C function by Paul Bourke 
 
/*
   Return the clockwise status of a curve, clockwise or counterclockwise
   n vertices making up curve p
   return 0 for incomputables eg: colinear points
          CLOCKWISE == 1
          COUNTERCLOCKWISE == -1
   It is assumed that
   - the polygon is closed
   - the last point is not repeated.
   - the polygon is simple (does not intersect itself or have holes)
*/
int ClockWise(XY *p,int n)
{
   int i,j,k;
   int count = 0;
   double z;
 
   if (n < 3)
      return(0);
 
   for (i=0;i<n;i++) {
      j = (i + 1) % n;
      k = (i + 2) % n;
      z  = (p[j].x - p[i].x) * (p[k].y - p[j].y);
      z -= (p[j].y - p[i].y) * (p[k].x - p[j].x);
      if (z < 0)
         count--;
      else if (z > 0)
         count++;
   }
   if (count > 0)
      return(COUNTERCLOCKWISE);
   else if (count < 0)
      return(CLOCKWISE);
   else
      return(0);
}
But I don't remember much C
I have tried to rewrite to C#:
private static int ClockWise(double[] y, double[]x, int n)
			
		{
			int i,j,k;
			int count = 0;
			double z;
 
			if (n < 3)//not a polygon
				return(0);
 
			for (i=0;i<n-1;i++) 
			{
				j = (i + 1) % n;
				k = (i + 2) % n;
				z  = (x[j] - x[i]) * (y[k] - y[j]);
				z -= (y[j] - y[i]) * (x[k] - x[j]);
				if (z < 0)
					count--;
				else if (z > 0)
					count++;
			}
			if (count > 0)//counterclockwise
				return(1);
			else if (count < 0)//clockwise
				return(-1);
			else
				return(0);
		}

Open in new window

0
Comment
Question by:MichaelMullin
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
6 Comments
 
LVL 26

Expert Comment

by:Anurag Thakur
ID: 24015899
try this one out
what i have tried here is creating a point class and using a generic list
i am assuming that you C program i taking the input of points array which is similar to the List<Point>
class Point
{
	public double x;
	public double y;
}
 
private int ClockWise (List<Point> p, int n)
{
	int i, j, k;
	int count = 0;
	double z;
 
	if (n < 3)
		return (0);
 
	for (i = 0; i < n; i++)
	{
		j = (i + 1) % n;
		k = (i + 2) % n;
		z = (p[j].x - p[i].x) * (p[k].y - p[j].y);
		z -= (p[j].y - p[i].y) * (p[k].x - p[j].x);
		if (z < 0)
			count--;
		else if (z > 0)
			count++;
	}
	if (count > 0)
		return (-1);
	else if (count < 0)
		return (1);
	else
		return (0);
}

Open in new window

0
 

Author Comment

by:MichaelMullin
ID: 24018068
Thank-you,
No, I do not use C. The question is: did I properly translate the C code to C#? My C# translation works for a simple 4-sided polygon but not for a 2500-sided polygon. Either the original C code is incorrect or my translation is incorrect. The simplest solution would be that my translation is incorrect. If you are able to assure me that my translation is either correct or incorrect, that would be the solution to this question.
0
 

Accepted Solution

by:
MichaelMullin earned 0 total points
ID: 24023176
I have solved this problem, and would like to close this question. I did not actually get an answer to my question, but than-you ragi0017: for you input. If someone still wants to answer the basic question, that is: Is my translation of the C code to C# correct or not, I am still interested. If the translation is correct, the C code algorithm is incorrect (or not robust enough for my needs) I did find another alorithm which I was able to adapt to C# and it works.
0
Concerto's Cloud Advisory Services

Want to avoid the missteps to gaining all the benefits of the cloud? Learn more about the different assessment options from our Cloud Advisory team.

 
LVL 26

Expert Comment

by:Anurag Thakur
ID: 24025306
no problems you can close the question
0
 

Author Comment

by:MichaelMullin
ID: 24025386
Even though I have solved this problem, I would still like to know if my translation from C to C# is correct or not. I will award the points for a yes or no answer with explanation if incorrect.
0
 
LVL 4

Expert Comment

by:VIkasumit
ID: 30720329
Although it is irrelevant, but for those who too want to check if conversion is good or not, then with my knowledge it is good.
0

Featured Post

Free Tool: IP Lookup

Get more info about an IP address or domain name, such as organization, abuse contacts and geolocation.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This article is meant to give a basic understanding of how to use R Sweave as a way to merge LaTeX and R code seamlessly into one presentable document.
Real-time is more about the business, not the technology. In day-to-day life, to make real-time decisions like buying or investing, business needs the latest information(e.g. Gold Rate/Stock Rate). Unlike traditional days, you need not wait for a fe…
This theoretical tutorial explains exceptions, reasons for exceptions, different categories of exception and exception hierarchy.
The viewer will be introduced to the member functions push_back and pop_back of the vector class. The video will teach the difference between the two as well as how to use each one along with its functionality.
Suggested Courses

604 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question