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Moving Cursor, using Pythagorous - Nietod

Posted on 2000-04-10
Medium Priority
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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
Question by:AntBon
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Author Comment

ID: 2700683
Adjusted points from 10 to 25
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LVL 22

Accepted Solution

nietod earned 100 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

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

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