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 Character movement with Mouse  (Read 1669 times) 0 Members and 1 Guest are viewing this topic.
vyh

Senior Newbie

 « Posted 2011-07-23 21:22:39 »

Hello,
Sorry for newbie question but How to movement Game Character with Mouse?
Something like I am have: mouseX, mouseY, playerX = 0, playerY = 0.
If I am click mouse get cordinates in mouseX and mouseY(Something like now mouseX = 5 and mouseY = 10) and move Character to x 1 and y 3(playerX = 1 and playerY = 3 now)
How to move Game character(wihout Teleport) to mouseX and mouseY cordinates?

Simple code snippets very useful.

Thanks and sorry for my English language.
Karmington

Senior Devvie

Medals: 1
Projects: 1

Co-op Freak

 « Reply #1 - Posted 2011-07-23 21:36:53 »

movement over time: http://en.wikipedia.org/wiki/Linear_interpolation

The simplest case, two known points. A good exercise to code yourself, though a googlefu may find it readyrolled.

Dx4

Junior Devvie

Medals: 5

 « Reply #2 - Posted 2011-07-24 07:40:32 »

first you need to find the angle

 1 `double angle = Math.atan2(mouseY-playerY,mouseX-playerX);`

then you need to build a vector to calculate the movement threshold

 1  2 `double movex = Math.cos(angle);double movey = Math.sin(angle);`

and now to move towards the point

 1  2  3  4  5 `x += movex;y += movey;if (x == mouseX && y == mouseY) { // you can use epsilon here stop();}`

hth
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theagentd

« JGO Bitwise Duke »

Medals: 409
Projects: 2
Exp: 8 years

 « Reply #3 - Posted 2011-07-24 12:20:01 »

and now to move towards the point

 1  2  3  4  5 `x += movex;y += movey;if (x == mouseX && y == mouseY) { // you can use epsilon here stop();}`

hth
That is NOT going to work. In 99.999999999999% of all cases it will just overshoot and continue forever. You would need to change the if-statement to detect overshooting and then set the position to (mouseX, mouseY) to actually land on the mouse position.

Myomyomyo.
Dx4

Junior Devvie

Medals: 5

 « Reply #4 - Posted 2011-07-24 13:03:04 »

and now to move towards the point

 1  2  3  4  5 `x += movex;y += movey;if (x == mouseX && y == mouseY) { // you can use epsilon here stop();}`

hth
That is NOT going to work. In 99.999999999999% of all cases it will just overshoot and continue forever. You would need to change the if-statement to detect overshooting and then set the position to (mouseX, mouseY) to actually land on the mouse position.

thats why I also write use an epsilon here in the comment, its up to OP to implement an epsilon though.

- David
h3ckboy

JGO Coder

Medals: 5

 « Reply #5 - Posted 2011-07-24 13:56:52 »

just quick question, is epsilon sort of like margin of error? so it can be off slightly of the exact point, and doesnt skip it?
dishmoth
 « Reply #6 - Posted 2011-07-24 16:27:20 »

thats why I also write use an epsilon here in the comment, its up to OP to implement an epsilon though.

I think theagentd's point is that there's no guarantee that N steps, for any integer N, will land the player on the target position, either exactly or within the epsilon margin-of-error.  You'd need to choose the step length (as well as the step direction) carefully to achieve that.  Or (as Karmington said) interpolate between the original position and the target position.  Or do something like the following, which moves the player towards the target at a fixed speed.

 1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20 `static final double stepSize = 5.0;  // the player's speed (how far to move each update)double playerX, playerY; // current positiondouble targetX, targetY; // where the mouse click happened// move the player one step closer to the target positionvoid updatePosition() {  double dx = targetX - playerX,         dy = targetY - playerY;  double distance = Math.hypot(dx,dy);  if ( distance > stepSize ) {    // still got a way to go, so take a full step    playerX += stepSize*dx/distance;    playerY += stepSize*dy/distance;  } else {    // we're either at the target already, or within one step of it    playerX = targetX;    playerY = targetY;  }}`

Simon

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