How could I make an optimized canvas pixel rendering? - javascript

I'm researching for a method for drawing a pixel with a well optimised method,
reducing the probability of fps drops. I use 2d, because it's more easy than webgl contexts. Here is a code that I tried:
<script>
document.write(".")
document.body.innerHTML=""
document.body.style.margin="0 0 0 0"
c=document.createElement("canvas")
document.body.appendChild(c)
ctx=c.getContext("2d")
setInterval(function(){
c.width=innerWidth
c.height=innerHeight
for(x=0;x<innerWidth;x++){
for(y=0;y<innerHeight;y++){
ctx.fillStyle="rgba(0,0,0,1)"
ctx.fillRect(x,y,x+1,y+1)
}}},1)
</script>
But when I save this as html file and open the html file with a navigator, the navigator gets caught and takes more than 2 seconds for load an image.

(() => {
document.write(".")
document.body.innerHTML=""
document.body.style.margin="0 0 0 0"
c=document.createElement("canvas")
document.body.appendChild(c)
ctx=c.getContext("2d")
c.width=innerWidth
c.height=innerHeight
x = 0;
y = 0;
function drawPixel() {
ctx.fillStyle="rgba(0,0,0,1)";
ctx.fillRect(x,y,x+1,y+1);
x ++;
if (x > innerWidth) {
x = 0;
y ++;
}
if (y < innerHeight)
requestAnimationFrame(drawPixel);
}
requestAnimationFrame(drawPixel)
})()
Try this, it's almost the same you did. You need to change some parts to get a better perfomance, i just adapted your code to work with requestAnimationFrame.

The quickest way to draw pixels one at a time is to create a image data object and add a typed array that holds each pixel as a 32 bit integer.
// assuming the ctx has been defined
var pixelData = ctx.getImageData(0, 0, ctx.canvas.width, ctx.canvas.height);
pixelData.data32 = new Uint32Array(pixelData.data.buffer); // create 32bit pixel
The quickest way to fill the buffered pixel data is via the typedArray.fill function
const white = 0xFFFFFFFF;
pixelData.data32.fill(white);
Or green
const green = 0xFF00FF00;
pixelData.data32.fill(green);
To set on pixel via its coordinates
function setPixel(x,y,color){
pixelData.data32[x + y * ctx.canvas.width] = color;
}
setPixel(10,10,0xFFFFFFFF);
To move the pixels to the canvas use
ctx.putImageData(pixelData,0,0);
The colour channels for a 32bit pixels are in the order ABGR such that 0xFF000000 is black 0x00000000 is transparent, 0xFFFF0000 is blue, 0xFF00FF00 is green and 0xFF0000FF is red. (note in some very rare situations the can change depending on the hardware's Endianness )

Related

Why does my canvas noise function always only appear red?

I'm trying to apply a noise effect to my canvas, based on a codepen I saw, which in turn appears to be very similar to an SO answer.
I want to produce a "screen" of randomly transparent pixels, but instead of that I get a field that's completely opaque red. I'm hoping someone who is more familiar with either canvas or typed arrays can show me what I'm doing wrong, and maybe help me understand a few of the techniques at play.
I refactored the codepen code significantly, because (for now) I don't care about animating the noise:
/**
* apply a "noise filter" to a rectangular region of the canvas
* #param {Canvas2DContext} ctx - the context to draw on
* #param {Number} x - the x-coordinate of the top-left corner of the region to noisify
* #param {Number} y - the y-coordinate of the top-left corner of the region to noisify
* #param {Number} width - how wide, in canvas units, the noisy region should be
* #param {Number} height - how tall, in canvas units, the noisy region should be
* #effect draws directly to the canvas
*/
function drawNoise( ctx, x, y, width, height ) {
let imageData = ctx.createImageData(width, height)
let buffer32 = new Uint32Array(imageData.data.buffer)
for (let i = 0, len = buffer32.length; i < len; i++) {
buffer32[i] = Math.random() < 0.5
? 0x00000088 // "noise" pixel
: 0x00000000 // non-noise pixel
}
ctx.putImageData(imageData, x, y)
}
From what I can tell, the core of what's happening is that we wrap the ImageData's raw data representation (a series of 8-bit elements that reflect the red, green, blue, and alpha values for each pixel, in series) in a 32-bit array, which allows us to operate on each pixel as a united tuple. We get an array with one element per pixel instead of four elements per pixel.
Then, we iterate through the elements in that array, writing RGBA values to each element (i.e. each pixel) based on our noise logic. The noise logic here is really simple: each pixel has a ~50% chance of being a "noise" pixel.
Noise pixels are assigned the 32-bit value 0x00000088, which (thanks to the 32-bit chunking provided by the array) is equivalent to rgba(0, 0, 0, 0.5), i.e. black, 50% opacity.
Non-noise pixels are assigned the 32-bit value 0x00000000, which is black 0% opacity, i.e. completely transparent.
Interestingly, we don't write the buffer32 to the canvas. Instead, we write the imageData that was used to construct the Uint32Array, leading me to believe that we're mutating the imageData object through some kind of pass-by-reference; I'm not clear exactly why this is. I know how value & reference passing works generally in JS (scalars are passed by value, objects are passed by reference), but in the non-typed array world, the value passed to the array constructor just determines the length of the array. That's evidently not what's happening here.
As noted, instead of a field of black pixels that are either 50% or 100% transparent, I get a field of all solid pixels, all red. Not only do I not expect to see the color red, there's zero evidence of the random color assignment: every pixel is solid red.
By playing with the two hex values, I've discovered that this produces a scattering of red on black that has the right kind of distribution:
buffer32[i] = Math.random() < 0.5
? 0xff0000ff // <-- I'd assume this is solid red
: 0xff000000 // <-- I'd assume this is invisible red
But it's still solid red, on solid black. None of the underlying canvas data shows through the pixels that should be invisible.
Confusingly, I can't get any colors other than red or black. I also can't get any transparency other than 100% opaque. Just to illustrate the disconnect, I've removed the random element and tried writing each of these nine values to every pixel just to see what happens:
buffer32[i] = 0xRrGgBbAa
// EXPECTED // ACTUAL
buffer32[i] = 0xff0000ff // red 100% // red 100%
buffer32[i] = 0x00ff00ff // green 100% // red 100%
buffer32[i] = 0x0000ffff // blue 100% // red 100%
buffer32[i] = 0xff000088 // red 50% // blood red; could be red on black at 50%
buffer32[i] = 0x00ff0088 // green 50% // red 100%
buffer32[i] = 0x0000ff88 // blue 50% // red 100%
buffer32[i] = 0xff000000 // red 0% // black 100%
buffer32[i] = 0x00ff0000 // green 0% // red 100%
buffer32[i] = 0x0000ff00 // blue 0% // red 100%
What's going on?
EDIT: similar (bad) results after dispensing with the Uint32Array and the spooky mutation, based on the MDN article on ImageData.data:
/**
* fails in exactly the same way
*/
function drawNoise( ctx, x, y, width, height ) {
let imageData = ctx.createImageData(width, height)
for (let i = 0, len = imageData.data.length; i < len; i += 4) {
imageData.data[i + 0] = 0
imageData.data[i + 1] = 0
imageData.data[i + 2] = 0
imageData.data[i + 3] = Math.random() < 0.5 ? 255 : 0
}
ctx.putImageData(imageData, x, y)
}
[TLDR]:
Your Hardware's endianness is designed as LittleEndian and thus the correct Hex format is 0xAABBGGRR, not 0xRRGGBBAA.
First let's explain the "magic" behind TypedArrays: ArrayBuffers.
An ArrayBuffer is a very special object which is directly linked to the device's memory. In itself the ArrayBuffer interface doesn't have too much features for us, but when you create one, you actually allocated its length in memory, for your own script. That is, the js engine won't deal with reallocating it, moving it somewhere else, chunking it and all these slow operations like it does with usual JS objects.
This thus makes it one of the fastest objects to manipulate binary data.
However, as said before, its interface is in itself quite limited. We have no way to access the data directly from the ArrayBuffer, to do this we have to use a view object, which won't copy the data, but really just offer a mean to access it directly.
You can have different views over the same ArrayBuffer, but the data used will always just be the one of the ArrayBuffer, and if you do edit an ArrayBuffer from one view, then it will be visible from the other:
const buffer = new ArrayBuffer(4);
const view1 = new Uint8Array(buffer);
const view2 = new Uint8Array(buffer);
console.log('view1', ...view1); // [0,0,0,0]
console.log('view2', ...view2); // [0,0,0,0]
// we modify only view1
view1[2] = 125;
console.log('view1', ...view1); // [0,0,125,0]
console.log('view2', ...view2); // [0,0,125,0]
There are different kind of view objects, and each will offer different ways to represent the binary data that is assigned to the memory slot allocated by the ArrayBuffer.
TypedArrays like Uint8Array, Float32Array etc. are ArrayLike interfaces which offer an easy way to manipulate the data as an Array, representing the data in their own format (8bits, Float32 etc.).
The DataView interface allows for more open manipulations like reading in different formats even from normally invalid boundaries, however, it comes at the cost of performance.
The ImageData interface itself uses an ArrayBuffer to store its pixel data. By default, it exposes an Uint8ClampedArray view over this data. That is, an ArrayLike object, with each 32bits pixel represented as values from 0 to 255 for each channel Red, Green, Blue and Alpha, in this order.
So your code is taking advantage of the fact TypedArrays are only view objects and that having an other view over the underlying ArrayBuffer will modify it directly.
Its author chose to use an Uint32Array because its a way to set a full pixel (remember canvas image is 32bits) in a single shot. You can reduce the work needed by four time.
However, doing so, you start dealing with 32bits values. And this may come a bit problematic, because now endianness matters.
The Uint8Array [0x00, 0x11, 0x22, 0x33] will be represented as the 32bits value 0x00112233 in BigEndian systems, but as 0x33221100 in LittleEndian ones.
const buff = new ArrayBuffer(4);
const uint8 = new Uint8Array(buff);
const uint32 = new Uint32Array(buff);
uint8[0] = 0x00;
uint8[1] = 0x11;
uint8[2] = 0x22;
uint8[3] = 0x33;
const hex32 = uint32[0].toString(16);
console.log(hex32, hex32 === "33221100" ? 'LE' : 'BE');
Note that most personal hardware are LittleEndian, so it's no surprise if your computer also is.
So with all this, I hope you do know how to fix your code: to generate the color rgba(0,0,0,.5), you need to set the Uint32 value 0x80000000
drawNoise(canvas.getContext('2d'), 0, 0, 300, 150);
function drawNoise(ctx, x, y, width, height) {
const imageData = ctx.createImageData(width, height)
const buffer32 = new Uint32Array(imageData.data.buffer)
const LE = isLittleEndian();
// 0xAABBRRGG : 0xRRGGBBAA;
const black = LE ? 0x80000000 : 0x00000080;
const blue = LE ? 0xFFFF0000 : 0x0000FFFF;
for (let i = 0, len = buffer32.length; i < len; i++) {
buffer32[i] = Math.random() < 0.5
? black
: blue
}
ctx.putImageData(imageData, x, y)
}
function isLittleEndian() {
const uint8 = new Uint8Array(8);
const uint32 = new Uint32Array(uint8.buffer);
uint8[0] = 255;
return uint32[0] === 0XFF;
}
<canvas id="canvas"></canvas>

Zoom my drawing on the background [duplicate]

This question already has an answer here:
HTML5 canvas zoom where mouse coordinates
(1 answer)
Closed 8 years ago.
I make program like a paint with HTML5 canvas and javascript. Drawing takes place on the background image. How to zoom my drawing on the background together.
Before zoom it:
After zoom it (need this result):
Note: zoom should be where clicked with the mouse on the background image
I've done this before!
First of all, I set a zoom level attribute on my canvas.
Main.canvas.zoomX = 1;
Main.canvas.zoomY = 1;
I also retain the original size of the canvas for reference.
Main.canvas.originW = Main.canvas.width;
Main.canvas.originH = Main.canvas.height;
I also retain the original left and top of the canvas for reference.
Main.canvas.gLeftStart = 0;
Main.canvas.gTopStart = 0;
I then set a zoom percentage. The zoom level will be adjusted by this amount every time that the zoom event occurs.
Main.canvas.zoomPerc = 0.05;
Next, I set an event listener on my canvas to watch for mousewheel.
Main.canvas.addEventListener('wheel', zoom, true);
Now, I'm going to write a quick function to retrieve the zoom, then I'll explain it.
function zoom(evt)
{
var x;
var y;
Main.canvas.xLayerS = (evt.layerX + (Main.canvas.gLeftStart * -1)) / (Main.canvas.originW * Main.canvas.zoomX);
Main.canvas.yLayerS = (evt.layerY + (Main.canvas.gTopStart * -1)) / (Main.canvas.originH * Main.canvas.zoomY);
Main.canvas.leftPerc = Main.canvas.gLeftStart / (Main.canvas.originW * Main.canvas.zoomX);
Main.canvas.topPerc = Main.canvas.gTopStart / (Main.canvas.originH * Main.canvas.zoomY);
if(evt.deltaY > 1)
{
Main.canvas.zoomX *= 1 + Main.canvas.zoomPerc;
Main.canvas.zoomY *= 1 + Main.canvas.zoomPerc;
}
else
{
Main.canvas.zoomX *= 1 - Main.canvas.zoomPerc;
Main.canvas.zoomY *= 1 - Main.canvas.zoomPerc;
}
var iiDS;
var cmd;
Main.canvas.xLayer = Main.canvas.xLayerS * (Main.canvas.originW * Main.canvas.zoomX);
Main.canvas.yLayer = Main.canvas.yLayerS * (Main.canvas.originH * Main.canvas.zoomY);
Main.context.clearRect(0, 0, Main.canvas.width, Main.canvas.height);
Main.context.beginPath();
Main.canvas.gLeftStart = (evt.layerX - Main.canvas.xLayer);
Main.canvas.gTopStart = (evt.layerY - Main.canvas.yLayer);
for(iiDS = 0; iiDS < Main.dataPoints.length; iiDS++)
{
if(iiDS === 0)
{
cmd = 'moveTo';
}
else
{
cmd = 'lineTo';
}
Main.dataPoints[iiDS].xPerc = Main.dataPoints[iiDS].x / Main.range.x;
Main.dataPoints[iiDS].yPerc = Main.dataPoints[iiDS].y / Main.range.y;
x = Main.canvas.gLeftStart + (Main.dataPoints[iiDS].xPerc * (Main.canvas.originW * Main.canvas.zoomX));
y = Main.canvas.gTopStart + (Main.dataPoints[iiDS].yPerc * (Main.canvas.originH * Main.canvas.zoomY));
Main.context[cmd](x, y);
}
Main.context.stroke();
}
Now that your canvas has been re-sized, you will need to redraw whatever was in it. Remember, any time that you re-size a canvas, you clear the canvas. If your canvas was holding an image, then that's simple, redraw that image at that size. If you canvas was holding data points (like a chart) then I would suggest that you make your data points have percentage like (probably a word for that) positions along your chart, not pixel positions.
More importantly though, I do not suggest that you ever re-size and re-position your canvas on zoom. Your page can get jumbled up and sloppy that way. Instead, use the percentages for size (like I showed you) and use the values for left and top positioning as starting points in your drawing. If a data point was a certain percentage of a way across a chart, it can be drawn at any size. Plus, you can draw outside of your canvas, it just won't be visible. Your canvas would then be more like a view-port.
You can do some really impressive charting this way, which a lot of companies pay a lot of money for. Have fun!
Did you try Context2d.scale(x, y)? You could do the following
var canvas = document.getElementById('myCanvas');
var context = canvas.getContext('2d');
context.scale(2, 2);
paintBackGround(context);
paintForeGround(context);
scale(factorWidth, factorHeight) Scales all coordinates in the canvas by the factors, so it will scale the background and the drawing. The example would double the size. You don't have to scale your coordinates by yourself, just let canvas do that for you.
Here is an example :
http://www.html5canvastutorials.com/advanced/html5-canvas-transform-scale-tutorial/
The only problem here: you need to scale before you draw, so you need a model that contains the original drawing in original unscaled coordinates, that can be drawn after scaling (paintForeGround() in my example)
Scale() is part of so called Transformations. You can Translate (move along a vector) rotate and scale the content of a canvas by using buildin functions of canvas. Just take a look at the html5canvastutorials. This works with matrix-mutliplications in the background, but it is really simple to use.

mouse position to isometric tile including height

Struggeling translating the position of the mouse to the location of the tiles in my grid. When it's all flat, the math looks like this:
this.position.x = Math.floor(((pos.y - 240) / 24) + ((pos.x - 320) / 48));
this.position.y = Math.floor(((pos.y - 240) / 24) - ((pos.x - 320) / 48));
where pos.x and pos.y are the position of the mouse, 240 and 320 are the offset, 24 and 48 the size of the tile. Position then contains the grid coordinate of the tile I'm hovering over. This works reasonably well on a flat surface.
Now I'm adding height, which the math does not take into account.
This grid is a 2D grid containing noise, that's being translated to height and tile type. Height is really just an adjustment to the 'Y' position of the tile, so it's possible for two tiles to be drawn in the same spot.
I don't know how to determine which tile I'm hovering over.
edit:
Made some headway... Before, I was depending on the mouseover event to calculate grid position. I just changed this to do the calculation in the draw loop itself, and check if the coordinates are within the limits of the tile currently being drawn. creates some overhead tho, not sure if I'm super happy with it but I'll confirm if it works.
edit 2018:
I have no answer, but since this ha[sd] an open bounty, help yourself to some code and a demo
The grid itself is, simplified;
let grid = [[10,15],[12,23]];
which leads to a drawing like:
for (var i = 0; i < grid.length; i++) {
for (var j = 0; j < grid[0].length; j++) {
let x = (j - i) * resourceWidth;
let y = ((i + j) * resourceHeight) + (grid[i][j] * -resourceHeight);
// the "+" bit is the adjustment for height according to perlin noise values
}
}
edit post-bounty:
See GIF. The accepted answer works. The delay is my fault, the screen doesn't update on mousemove (yet) and the frame rate is low-ish. It's clearly bringing back the right tile.
Source
Intresting task.
Lets try to simplify it - lets resolve this concrete case
Solution
Working version is here: https://github.com/amuzalevskiy/perlin-landscape (changes https://github.com/jorgt/perlin-landscape/pull/1 )
Explanation
First what came into mind is:
Just two steps:
find an vertical column, which matches some set of tiles
iterate tiles in set from bottom to top, checking if cursor is placed lower than top line
Step 1
We need two functions here:
Detects column:
function getColumn(mouseX, firstTileXShiftAtScreen, columnWidth) {
return (mouseX - firstTileXShiftAtScreen) / columnWidth;
}
Function which extracts an array of tiles which correspond to this column.
Rotate image 45 deg in mind. The red numbers are columnNo. 3 column is highlighted. X axis is horizontal
function tileExists(x, y, width, height) {
return x >= 0 & y >= 0 & x < width & y < height;
}
function getTilesInColumn(columnNo, width, height) {
let startTileX = 0, startTileY = 0;
let xShift = true;
for (let i = 0; i < columnNo; i++) {
if (tileExists(startTileX + 1, startTileY, width, height)) {
startTileX++;
} else {
if (xShift) {
xShift = false;
} else {
startTileY++;
}
}
}
let tilesInColumn = [];
while(tileExists(startTileX, startTileY, width, height)) {
tilesInColumn.push({x: startTileX, y: startTileY, isLeft: xShift});
if (xShift) {
startTileX--;
} else {
startTileY++;
}
xShift = !xShift;
}
return tilesInColumn;
}
Step 2
A list of tiles to check is ready. Now for each tile we need to find a top line. Also we have two types of tiles: left and right. We already stored this info during building matching tiles set.
function getTileYIncrementByTileZ(tileZ) {
// implement here
return 0;
}
function findExactTile(mouseX, mouseY, tilesInColumn, tiles2d,
firstTileXShiftAtScreen, firstTileYShiftAtScreenAt0Height,
tileWidth, tileHeight) {
// we built a set of tiles where bottom ones come first
// iterate tiles from bottom to top
for(var i = 0; i < tilesInColumn; i++) {
let tileInfo = tilesInColumn[i];
let lineAB = findABForTopLineOfTile(tileInfo.x, tileInfo.y, tiles2d[tileInfo.x][tileInfo.y],
tileInfo.isLeft, tileWidth, tileHeight);
if ((mouseY - firstTileYShiftAtScreenAt0Height) >
(mouseX - firstTileXShiftAtScreen)*lineAB.a + lineAB.b) {
// WOHOO !!!
return tileInfo;
}
}
}
function findABForTopLineOfTile(tileX, tileY, tileZ, isLeftTopLine, tileWidth, tileHeight) {
// find a top line ~~~ a,b
// y = a * x + b;
let a = tileWidth / tileHeight;
if (isLeftTopLine) {
a = -a;
}
let b = isLeftTopLine ?
tileY * 2 * tileHeight :
- (tileX + 1) * 2 * tileHeight;
b -= getTileYIncrementByTileZ(tileZ);
return {a: a, b: b};
}
Please don't judge me as I am not posting any code. I am just suggesting an algorithm that can solve it without high memory usage.
The Algorithm:
Actually to determine which tile is on mouse hover we don't need to check all the tiles. At first we think the surface is 2D and find which tile the mouse pointer goes over with the formula OP posted. This is the farthest probable tile mouse cursor can point at this cursor position.
This tile can receive mouse pointer if it's at 0 height, by checking it's current height we can verify if this is really at the height to receive pointer, we mark it and move forward.
Then we find the next probable tile which is closer to the screen by incrementing or decrementing x,y grid values depending on the cursor position.
Then we keep on moving forward in a zigzag fashion until we reach a tile which cannot receive pointer even if it is at it's maximum height.
When we reach this point the last tile found that were at a height to receive pointer is the tile that we are looking for.
In this case we only checked 8 tiles to determine which tile is currently receiving pointer. This is very memory efficient in comparison to checking all the tiles present in the grid and yields faster result.
One way to solve this would be to follow the ray that goes from the clicked pixel on the screen into the map. For that, just determine the camera position in relation to the map and the direction it is looking at:
const camPos = {x: -5, y: -5, z: -5}
const camDirection = { x: 1, y:1, z:1}
The next step is to get the touch Position in the 3D world. In this certain perspective that is quite simple:
const touchPos = {
x: camPos.x + touch.x / Math.sqrt(2),
y: camPos.y - touch.x / Math.sqrt(2),
z: camPos.z - touch.y / Math.sqrt(2)
};
Now you just need to follow the ray into the layer (scale the directions so that they are smaller than one of your tiles dimensions):
for(let delta = 0; delta < 100; delta++){
const x = touchPos.x + camDirection.x * delta;
const y = touchPos.y + camDirection.y * delta;
const z = touchPos.z + camDirection.z * delta;
Now just take the tile at xz and check if y is smaller than its height;
const absX = ~~( x / 24 );
const absZ = ~~( z / 24 );
if(tiles[absX][absZ].height >= y){
// hanfle the over event
}
I had same situation on a game. first I tried with mathematics, but when I found that the clients wants to change the map type every day, I changed the solution with some graphical solution and pass it to the designer of the team. I captured the mouse position by listening the SVG elements click.
the main graphic directly used to capture and translate the mouse position to my required pixel.
https://blog.lavrton.com/hit-region-detection-for-html5-canvas-and-how-to-listen-to-click-events-on-canvas-shapes-815034d7e9f8
https://code.sololearn.com/Wq2bwzSxSnjl/#html
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).

Canvas getImageData() For optimal performance. To pull out all data or one at a time?

I need to scan through every pixel in a canvas image and do some fiddling with the colors etc. For optimal performance, should I grab all the data in one go and work on it through the array? Or should I call each pixel as I work on it.
So basically...
data = context.getImageData(x, y, height, width);
VS
data = context.getImageData(x, y, 1, 1); //in a loop height*width times.
You'll get much higher performances by grabbing the image all at once since :
a) a (contiguous) acces to an array is way faster than a function call.
b) especially when this function isa method of a DOM object having some overhead.
c) and there might be buffer refresh issues that might delay response (if canvas is
on sight... or not depending on double buffering implementation).
So go for a one-time grab.
I'll suggest you look into Javascript Typed Arrays to get the most of the
imageData result.
If i may quote myself, look at how you can handle pixels fast in this old post of mine
(look after 2) ):
Nice ellipse on a canvas?
(i quoted the relevant part below : )
You can get a UInt32Array view on your ImageData with :
var myGetImageData = myTempCanvas.getImageData(0,0,sizeX, sizeY);
var sourceBuffer32 = new Uint32Array(myGetImageData.data.buffer);
then sourceBuffer32[i] contains Red, Green, Blue, and transparency packed into one unsigned 32 bit int. Compare it to 0 to know if pixel is non-black ( != (0,0,0,0) )
OR you can be more precise with a Uint8Array view :
var myGetImageData = myTempCanvas.getImageData(0,0,sizeX, sizeY);
var sourceBuffer8 = new Uint8Array(myGetImageData.data.buffer);
If you deal only with shades of grey, then R=G=B, so watch for
sourceBuffer8[4*i]>Threshold
and you can set the i-th pixel to black in one time using the UInt32Array view :
sourceBuffer32[i]=0xff000000;
set to any color/alpha with :
sourceBuffer32[i]= (A<<24) | (B<<16) | (G<<8) | R ;
or just to any color :
sourceBuffer32[i]= 0xff000000 | (B<<16) | (G<<8) | R ;
(be sure R is rounded).
Listening to #Ken's comment, yes endianness can be an issue when you start fighting with bits 32 at a time.
Most computer are using little-endian, so RGBA becomes ABGR when dealing with them 32bits a once.
Since it is the vast majority of systems, if dealing with 32bit integer assume this is the case,
and you can -for compatibility- reverse your computation before writing the 32 bits results on Big endian systems.
Let me share those two functions :
function isLittleEndian() {
// from TooTallNate / endianness.js. https://gist.github.com/TooTallNate/4750953
var b = new ArrayBuffer(4);
var a = new Uint32Array(b);
var c = new Uint8Array(b);
a[0] = 0xdeadbeef;
if (c[0] == 0xef) { isLittleEndian = function() {return true }; return true; }
if (c[0] == 0xde) { isLittleEndian = function() {return false }; return false; }
throw new Error('unknown endianness');
}
function reverseUint32 (uint32) {
var s32 = new Uint32Array(4);
var s8 = new Uint8Array(s32.buffer);
var t32 = new Uint32Array(4);
var t8 = new Uint8Array(t32.buffer);
reverseUint32 = function (x) {
s32[0] = x;
t8[0] = s8[3];
t8[1] = s8[2];
t8[2] = s8[1];
t8[3] = s8[0];
return t32[0];
}
return reverseUint32(uint32);
};
Additionally to what GameAlchemist said, if you want to get or set all the colors of a pixel simultaneously, but you don't want to check endianness, you can use a DataView:
var data = context.getImageData(0, 0, canvas.width, canvas.height);
var view = new DataView(data.data.buffer);
// Read or set pixel (x,y) as #RRGGBBAA (big endian)
view.getUint32(4 * (x + y*canvas.width));
view.setUint32(4 * (x + y*canvas.width), 0xRRGGBBAA);
// Read or set pixel (x,y) as #AABBGGRR (little endian)
view.getUint32(4 * (x + y*canvas.width), true);
view.setUint32(4 * (x + y*canvas.width), 0xAABBGGRR, true);
// Save changes
ctx.putImageData(data, 0, 0);
It depends on what exactly you're doing, but I'd suggest grabbing it all at once, and then looping through it.
Grabbing it all at once is faster than grabbing it pixel by pixel, since searching through an array is a lot faster than searching through a canvas, once for each pixel.
If you're really in need of speed, look into web workers. You can set each one to grab a specific section of the canvas, and since they can run simultaneously, they'll make much better use out of your CPU.
getImageData() isn't really slow enough for you to notice the difference if you were to grab it all at once or individually, in my experiences using the function.

HTML canvas double buffering frame-rate issues

I have a full-screen canvas with 3 images drawn on it. When I resize the window, these images change position; however, it appears to be very glitchy, more so in Firefox.
I've been reading that double-buffering should resolve this issue, but I'm wondering how I would double buffer when the next position is unknown. That is to say, I cannot determine what should be buffered in the future, so how would this be possible?
Here is one source that seems doable, but I do not fully understand the concept Fedor is trying to explain.
Does HTML5/Canvas Support Double Buffering?
So far I have,
$canvas = $('#myclouds')[0];
$canvas_buffer = $('canvas')[0].insertAfter($canvas).css('visibility', 'hidden');
context = $canvas.getContext('2d');
context_buffer = $canvas_buffer.getContext('2d');
clouds_arr = [$canvas, $canvas_buffer];
$(window).resize(function () {
drawCanvas();
};
function initCanvas() {
// Sources for cloud images
var cloud1 = '/js/application/home/images/cloud1.png',
cloud2 = '/js/application/home/images/cloud2.png',
cloud3 = '/js/application/home/images/cloud3.png';
// add clouds to be drawn
// parameters are as follows:
// image source, x, y, ratio, adjustment)
addCloud(cloud1, null, 125, .03);
addCloud(cloud2, null, 75, .15);
addCloud(cloud3, null, 50, .55);
addCloud(cloud1, null, 125, .97, 300);
addCloud(cloud2, null, 70, .85, 300);
addCloud(cloud3, null, 45, .5, 300);
// Draw the canvas
drawCanvas();
}
function drawCanvas() {
// Reset
$canvas.attr('height', $window.height()).attr('width', $window.width());
// draw the clouds
var l = clouds.length;
for (var i = 0; i < l; i++) {
clouds[i].x = ($window.width() * clouds[i].ratio) - clouds[i].offset;
drawimage(context, clouds[i]);
}
}
function Cloud() {
this.x = 0;
this.y = 0;
}
function addCloud(path, x, y, ratio, offset) {
var c = new Cloud;
c.x = x;
c.y = y;
c.path = path;
c.ratio = ratio || 0;
c.offset = offset || 0;
clouds.push(c);
}
function drawimage(ctx, image) {
var clouds_obj = new Image();
clouds_obj.src = image.path;
clouds_obj.onload = function() {
ctx.drawImage(clouds_obj, image.x, image.y);
};
}
I think maybe you are misunderstanding what double buffering is. Its a technique for smooth real-time rendering of graphics on a display.
The concept is you have two buffers. Only one is visible at any one time. When you go to draw the elements that make up a frame you draw them to the invisible buffer. In you case the clouds. Then you flip the buffers making the hidden one visible and the visible one hidden. Then on the next frame you draw to the now newly hidden buffer. Then at the end of drawing you flip back.
What this does is stop the user seeing partial rendering of elements before a frame is complete. On gaming systems this would also be synced up with the vertical refresh of the display to be really smooth and stop artefacts such as tearing to occur.
Looking at you code above you seem to have created the two canvas elements, but you're only using the first Context object. I assume this is incomplete as no flipping is taking place.
Its also worth noting that the window resize event can fire continuously when dragging which can cause frantic rendering. I usually create a timer on the resize event to actually re-render. This way the re-render only happens once the user stops resizing for a few milliseconds.
Also, your draw routine is creating new Image objects every time which you don't need to do. You can use one image object and render to the canvas multiple times. This will speed up your render considerably.
Hope this helps.

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