Solved

Moving Cursor, using Pythagorous - Nietod

Posted on 2000-04-10
3
206 Views
Last Modified: 2010-04-02
nietod, you previously advised me to implement 'Pythagorous's theory' to move the cursor.

I have tried to do what you said, but i am still having problems..can you help me out - or tell me where i am going wrong

here is the code


if (Dest.x > Source.x) //Move to the right
   {
//   xdir = RIGHT;
   xlength=Dest.x - Source.x;
//   xdistance=Dest.x - Source.x;
   }
else
   {
   xlength = Source.x - Dest.x; //Move to the left
//   xdistance = Source.x - Dest.x; //Move to the left
//   xdir = LEFT;
   }

if (Dest.y > Source.y) //Move Downwards
   {
   ylength=Dest.y - Source.y;
//   ydistance=Dest.y - Source.y;
//   ydir = DOWN;
   }
else
   {
   ylength = Source.y - Dest.y; //Move upwards
//   ydistance = Source.y - Dest.y;
//   ydir = UP;
   }


   xdistance=xlength;
   ydistance=ylength;

   xlength=xlength*xlength;
   ylength=ylength*ylength;

   hypot=abs(sqrt(xlength+ylength)); // work out length of the hypotenuse in pixels


   for (i=0;i<=hypot;i++) // move in pixels
   {
   xpos=abs((Source.x+(i*((float)xdistance/hypot))));
   ypos=abs((Source.y+(i*((float)ydistance/hypot))));

   SetCursorPos(xpos,ypos);
   Application->ProcessMessages();
   for (int pause = 0;pause < 1000000; pause++);;
   }


thanks

PS i have got new delay mechanism :-)

thanks
0
Comment
Question by:AntBon
  • 2
3 Comments
 

Author Comment

by:AntBon
ID: 2700683
Adjusted points from 10 to 25
0
 
LVL 22

Accepted Solution

by:
nietod earned 25 total points
ID: 2700810
POINT Dst;
POINT Src;
int DeltaX  = Dst.x - Src.x;
int DeltaY  = Dst.y - Src.y;

// this is perhaps rounded down, that is fine.
 int distance = (int) sqrt(DeltaX*DeltaX + DeltaY*DeltaY);

// you may wnat to start i at 1, not 0,  starting at 0
// makes the agorithm incude the Source point.  
for (int i = 0; i <=1 distance; ++i)
{
   int x = Src.x+ i*DeltaX/distance;
   int y =  Src.y+i*DeltaY/distance;
   // move cursor or do something with x and y.
}

That is it.  No need to worry about signs.  DeltaX and DeltaY are signed.  Although some rounding occurs (+ or - 1) for the interior points (because of integer coordinates) The final value of x and y should be exactly the destination coordinates.  (i.e there is no rounding errors in the final coordinates.
0
 

Author Comment

by:AntBon
ID: 2700882
Thanks very much, but i have just done it myself, but at least i can compare my solution with your, on the face of it they look the same, except i started my for loop at 0, but i will change now


Thanks Again
0

Featured Post

Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

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

Suggested Solutions

What is C++ STL?: STL stands for Standard Template Library and is a part of standard C++ libraries. It contains many useful data structures (containers) and algorithms, which can spare you a lot of the time. Today we will look at the STL Vector. …
This article will show you some of the more useful Standard Template Library (STL) algorithms through the use of working examples.  You will learn about how these algorithms fit into the STL architecture, how they work with STL containers, and why t…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.
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.

856 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