Solved

Moving Cursor, using Pythagorous - Nietod

Posted on 2000-04-10
3
209 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
[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
  • 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

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

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

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. …
Go is an acronym of golang, is a programming language developed Google in 2007. Go is a new language that is mostly in the C family, with significant input from Pascal/Modula/Oberon family. Hence Go arisen as low-level language with fast compilation…
The viewer will learn how to use the return statement in functions in C++. The video will also teach the user how to pass data to a function and have the function return data back for further processing.
The viewer will learn how to clear a vector as well as how to detect empty vectors in C++.

751 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