Defining boundaries of a hitbox in Javascript - javascript
I have been working on a simple brick breaker game in Javascript for the past few weeks and come across an issue with creating a hitbox that I have no clue on where to begin fixing it.
A brief overview of what's going on:
My Hitbox class takes in a gameObject (a ball, or a brick, etc; defined elsewhere) and in the constructor sets its own position, width and height as the given game object's respectively. It has 2 arrays that I use for collision detection, one for the left/right sides and one for the top/bottom. Lastly it has 4 objects (or in this case positions) that define the "side" of the hitbox. These objects contain a singular value and an array of values.
Since every hitbox is going to be some form of a rectangle, I can simplify the creation of a hitbox because both left/right sides are vertical lines (all x values will be the same) and both top/bottom are horizontal lines. (all y values will be the same)
For instance, starting with the left side of the hitbox:
I know the starting point will be the top left corner of the object (gameObject.position.y) and the ending point will be the bottom left corner. (gameObject.position.y + gameObject.height) So, all I need to do is loop through all the y values up to the ending point to create the left side of the hitbox. So, the positions of the left side of the hitbox would be:
(gameObject.position.x, sides[0])
(gameObject.position.x, sides[1])
(gameObject.position.x, sides[2])
...
(gameObject.position.x, sides[n])
And the same logic for the right side:
(gameObject.position.x + gameObject.width, sides[0])
(gameObject.position.x + gameObject.width, sides[1])
(gameObject.position.x + gameObject.width, sides[2])
...
(gameObject.position.x + gameObject.width, sides[n])
Here's the class itself. You can ignore the console.log() functions in the buildHitbox() member function; I was just using those for testing.
export default class Hitbox{
constructor(gameObject){
this.position = gameObject.position;
this.width = gameObject.width;
this.height = gameObject.height;
this.sides = [];
this.tops = [];
this.leftSide = {
x: gameObject.position.x,
y: this.sides
};
this.rightSide = {
x: gameObject.position.x + gameObject.width,
y: this.sides
};
this.topSide = {
x: this.tops,
y: gameObject.position.y
};
this.bottomSide = {
x: this.tops,
y: gameObject.position.y + gameObject.height
};
}
buildHitbox(){
// left/right sides of hitbox
for(var sideIndex = this.position.y; sideIndex < this.height; sideIndex++){
this.sides.push(sideIndex);
}
//console.log("hitbox left side: ", this.leftSide); //SUCCESSFULL
//console.log("hitbox right side: ", this.rightSide); //SUCCESSFULL
// top/bottom sides of hitbox
for(var topIndex = this.position.x; topIndex < this.width; topIndex++){
this.tops.push(topIndex);
}
//works for far left brick but not far right
//console.log("hitbox tops sides: ", this.topSide);
//console.log("hitbox bottom sides: ", this.bottomSide);
//works for far left brick but not for right
}
Here's my issue
The test case I'm using for this is a brick at the top left corner of the canvas and at the top right corner of the canvas. The canvas itself is 1000px x 600px, so the left brick will start at position (0,0) and the right brick will start at (950,0) because each bricks' width is 50px.
The left and right sides of the hitbox are working correctly, (seen below)
values of left/right side arrays
but when I apply the same logic to the top and bottom array, it works as intended for the left brick but doesn't populate the array for the right brick.
top/bottom values showing empty arrays
I have already checked to make sure that the right brick shows it's starting x position as 950, and it does so I know it's not an issue with my level generator.
Maybe it has something to do with the bounds of the canvas?
Maybe it has something to do with the IDE I'm using? (codesandbox.io)
Maybe it has something to do with how things are rendered in javascript?
Any help with this would be greatly appreciated and I am willing to share more aspects of the game if needed. I am just completely stuck at the moment and can't really progress until I figure this one out.
It is not anything else but the code bug.
You should use topIndex < this.position.x + this.width instead of topIndex < this.width.
The right code block is here:
for(var sideIndex = this.position.y; sideIndex < this.position.y + this.height; sideIndex++){
this.sides.push(sideIndex);
}
console.log("hitbox left side: ", this.leftSide); //SUCCESSFULL
console.log("hitbox right side: ", this.rightSide); //SUCCESSFULL
// top/bottom sides of hitbox
for(var topIndex = this.position.x; topIndex < this.position.x + this.width; topIndex++){
this.tops.push(topIndex);
}
//works for far left brick but not far right
console.log("hitbox tops sides: ", this.topSide);
console.log("hitbox bottom sides: ", this.bottomSide);
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mouse position to isometric tile including height
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Here is the grid input I would define for the sake of this discussion. The output should be some tile (coordinate_1, coordinate_2) based on visibility on the users screen of the mouse: I can offer two solutions from different perspectives, but you will need to convert this back into your problem domain. The first methodology is based on coloring tiles and can be more useful if the map is changing dynamically. The second solution is based on drawing coordinate bounding boxes based on the fact that tiles closer to the viewer like (0, 0) can never be occluded by tiles behind it (1,1). Approach 1: Transparently Colored Tiles The first approach is based on drawing and elaborated on here. I must give the credit to #haldagan for a particularly beautiful solution. In summary it relies on drawing a perfectly opaque layer on top of the original canvas and coloring every tile with a different color. This top layer should be subject to the same height transformations as the underlying layer. When the mouse hovers over a particular layer you can detect the color through canvas and thus the tile itself. This is the solution I would probably go with and this seems to be a not so rare issue in computer visualization and graphics (finding positions in a 3d isometric world). Approach 2: Finding the Bounding Tile This is based on the conjecture that the "front" row can never be occluded by "back" rows behind it. Furthermore, "closer to the screen" tiles cannot be occluded by tiles "farther from the screen". To make precise the meaning of "front", "back", "closer to the screen" and "farther from the screen", take a look at the following: . Based on this principle the approach is to build a set of polygons for each tile. So firstly we determine the coordinates on the canvas of just box (0, 0) after height scaling. Note that the height scale operation is simply a trapezoid stretched vertically based on height. Then we determine the coordinates on the canvas of boxes (1, 0), (0, 1), (1, 1) after height scaling (we would need to subtract anything from those polygons which overlap with the polygon (0, 0)). Proceed to build each boxes bounding coordinates by subtracting any occlusions from polygons closer to the screen, to eventually get coordinates of polygons for all boxes. With these coordinates and some care you can ultimately determine which tile is pointed to by a binary search style through overlapping polygons by searching through bottom rows up.
It also matters what else is on the screen. Maths attempts work if your tiles are pretty much uniform. However if you are displaying various objects and want the user to pick them, it is far easier to have a canvas-sized map of identifiers. function poly(ctx){var a=arguments;ctx.beginPath();ctx.moveTo(a[1],a[2]); for(var i=3;i<a.length;i+=2)ctx.lineTo(a[i],a[i+1]);ctx.closePath();ctx.fill();ctx.stroke();} function circle(ctx,x,y,r){ctx.beginPath();ctx.arc(x,y,r,0,2*Math.PI);ctx.fill();ctx.stroke();} function Tile(h,c,f){ var cnv=document.createElement("canvas");cnv.width=100;cnv.height=h; var ctx=cnv.getContext("2d");ctx.lineWidth=3;ctx.lineStyle="black"; ctx.fillStyle=c;poly(ctx,2,h-50,50,h-75,98,h-50,50,h-25); poly(ctx,50,h-25,2,h-50,2,h-25,50,h-2); poly(ctx,50,h-25,98,h-50,98,h-25,50,h-2); f(ctx);return ctx.getImageData(0,0,100,h); } function put(x,y,tile,image,id,map){ var iw=image.width,tw=tile.width,th=tile.height,bdat=image.data,fdat=tile.data; for(var i=0;i<tw;i++) for(var j=0;j<th;j++){ var ijtw4=(i+j*tw)*4,a=fdat[ijtw4+3]; if(a!==0){ var xiyjiw=x+i+(y+j)*iw; for(var k=0;k<3;k++)bdat[xiyjiw*4+k]=(bdat[xiyjiw*4+k]*(255-a)+fdat[ijtw4+k]*a)/255; bdat[xiyjiw*4+3]=255; map[xiyjiw]=id; } } } var cleanimage; var pickmap; function startup(){ var water=Tile(77,"blue",function(){}); var field=Tile(77,"lime",function(){}); var tree=Tile(200,"lime",function(ctx){ ctx.fillStyle="brown";poly(ctx,50,50,70,150,30,150); ctx.fillStyle="forestgreen";circle(ctx,60,40,30);circle(ctx,68,70,30);circle(ctx,32,60,30); }); var sheep=Tile(200,"lime",function(ctx){ ctx.fillStyle="white";poly(ctx,25,155,25,100);poly(ctx,75,155,75,100); circle(ctx,50,100,45);circle(ctx,50,80,30); poly(ctx,40,70,35,80);poly(ctx,60,70,65,80); }); var cnv=document.getElementById("scape"); cnv.width=500;cnv.height=400; var ctx=cnv.getContext("2d"); cleanimage=ctx.getImageData(0,0,500,400); pickmap=new Uint8Array(500*400); var tiles=[water,field,tree,sheep]; var map=[[[0,0],[1,1],[1,1],[1,1],[1,1]], [[0,0],[1,1],[1,2],[3,2],[1,1]], [[0,0],[1,1],[2,2],[3,2],[1,1]], [[0,0],[1,1],[1,1],[1,1],[1,1]], [[0,0],[0,0],[0,0],[0,0],[0,0]]]; for(var x=0;x<5;x++) for(var y=0;y<5;y++){ var desc=map[y][x],tile=tiles[desc[0]]; put(200+x*50-y*50,200+x*25+y*25-tile.height-desc[1]*20, tile,cleanimage,x+1+(y+1)*10,pickmap); } ctx.putImageData(cleanimage,0,0); } var mx,my,pick; function mmove(event){ mx=Math.round(event.offsetX); my=Math.round(event.offsetY); if(mx>=0 && my>=0 && mx<cleanimage.width && my<cleanimage.height && pick!==pickmap[mx+my*cleanimage.width]) requestAnimationFrame(redraw); } function redraw(){ pick=pickmap[mx+my*cleanimage.width]; document.getElementById("pick").innerHTML=pick; var ctx=document.getElementById("scape").getContext("2d"); ctx.putImageData(cleanimage,0,0); if(pick!==0){ var temp=ctx.getImageData(0,0,cleanimage.width,cleanimage.height); for(var i=0;i<pickmap.length;i++) if(pickmap[i]===pick) temp.data[i*4]=255; ctx.putImageData(temp,0,0); } } startup(); // in place of body.onload <div id="pick">Move around</div> <canvas id="scape" onmousemove="mmove(event)"></canvas> Here the "id" is a simple x+1+(y+1)*10 (so it is nice when displayed) and fits into a byte (Uint8Array), which could go up to 15x15 display grid already, and there are wider types available too. (Tried to draw it small, and it looked ok on the snippet editor screen but apparently it is still too large here)
Computer graphics is fun, right? This is a special case of the more standard computational geometry "point location problem". You could also express it as a nearest neighbour search. To make this look like a point location problem you just need to express your tiles as non-overlapping polygons in a 2D plane. If you want to keep your shapes in a 3D space (e.g. with a z buffer) this becomes the related "ray casting problem". One source of good geometry algorithms is W. Randolf Franklin's website and turf.js contains an implementation of his PNPOLY algorithm. For this special case we can be even faster than the general algorithms by treating our prior knowledge about the shape of the tiles as a coarse R-tree (a type of spatial index).
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