Related
Updated with more explanation around my confusion
(This is how a non-graphics developer imagines the rendering process!)
I specify a 2x2 sqaure to be drawn in by way of two triangles. I'm going to not talk about the triangle anymore. Square is a lot better. Let's say the square gets drawn in one piece.
I have not specified any units for my drawing. The only places in my code that I do something like that is: canvas size (set to 1x1 in my case) and the viewport (i always set this to the dimensions of my output texture).
Then I call draw().
What happens is this: that regardless of the size of my texture (being 1x1 or 10000x10000) all my texels are filled with data (color) that I returned from my frag shader. This is working each time perfectly.
So now I'm trying to explain this to myself:
The GPU is only concerned with coloring the pixels.
Pixel is the smallest unit that the GPU deals with (colors).
Depending on how many pixels my 2x2 square is mapped to, I should be running into one of the following 3 cases:
The number of pixels (to be colored) and my output texture dims match one to one: In this ideal case, for each pixel, there would be one value assigned to my output texture. Very clear to me.
The number of pixels are fewer than my output texture dims. In this case, I should expect that some of the output texels to have exact same value (which is the color of the pixel the fall under). For instance if the GPU ends up drawing 16x16 pixels and my texture is 64x64 then I'll have blocks of 4 texel which get the same value. I have not observed such case regardless of the size of my texture. Which means there is never a case where we end up with fewer pixels (really hard to imagine -- let's keep going)
The number of pixels end up being more than the number of texels. In this case, the GPU should decide which value to assign to my texel. Would it average out the pixel colors? If the GPU is coloring 64x64 pixels and my output texture is 16x16 then I should expect that each texel gets an average color of the 4x4 pixels it contains. Anyway, in this case my texture should be completely filled with values I didn't intend specifically for them (like averaged out) however this has not been the case.
I didn't even talk about how many times my frag shader gets called because it didn't matter. The results would be deterministic anyway.
So considering that I have never run into 2nd and 3rd case where the values in my texels are not what I expected them the only conclusion I can come up with is that the whole assumption of the GPU trying to render pixels is actually wrong. When I assign an output texture to it (which is supposed to stretch over my 2x2 square all the time) then the GPU will happily oblige and for each texel will call my frag shader. Somewhere along the line the pixels get colored too.
But the above lunatistic explanation also fails to answer why I end up with no values in my texels or incorrect values if I stretch my geometry to 1x1 or 4x4 instead of 2x2.
Hopefully the above fantastic narration of the GPU coloring process has given you clues as to where I'm getting this wrong.
Original Post:
We're using WebGL for general computation. As such we create a rectangle and draw 2 triangles in it. Ultimately what we want is the data inside the texture mapped to this geometry.
What I don't understand is if I change the rectangle from (-1,-1):(1,1) to say (-0.5,-0.5):(0.5,0.5) suddenly data is dropped from the texture bound to the framebuffer.
I'd appreciate if someone makes me understand the correlations. The only places that real dimensions of the output texture come into play are the call to viewPort() and readPixels().
Below are relevant pieces of code for you to see what I'm doing:
... // canvas is created with size: 1x1
... // context attributes passed to canvas.getContext()
contextAttributes = {
alpha: false,
depth: false,
antialias: false,
stencil: false,
preserveDrawingBuffer: false,
premultipliedAlpha: false,
failIfMajorPerformanceCaveat: true
};
... // default geometry
// Sets of x,y,z (for rectangle) and s,t coordinates (for texture)
return new Float32Array([
-1.0, 1.0, 0.0, 0.0, 1.0, // upper left
-1.0, -1.0, 0.0, 0.0, 0.0, // lower left
1.0, 1.0, 0.0, 1.0, 1.0, // upper right
1.0, -1.0, 0.0, 1.0, 0.0 // lower right
]);
...
const geometry = this.createDefaultGeometry();
gl.bindBuffer(gl.ARRAY_BUFFER, buffer);
gl.bufferData(gl.ARRAY_BUFFER, geometry, gl.STATIC_DRAW);
... // binding to the vertex shader attribs
gl.vertexAttribPointer(positionHandle, 3, gl.FLOAT, false, 20, 0);
gl.vertexAttribPointer(textureCoordHandle, 2, gl.FLOAT, false, 20, 12);
gl.enableVertexAttribArray(positionHandle);
gl.enableVertexAttribArray(textureCoordHandle);
... // setting up framebuffer; I set the viewport to output texture dimensions (I think this is absolutely needed but not sure)
gl.bindTexture(gl.TEXTURE_2D, texture);
gl.bindFramebuffer(gl.FRAMEBUFFER, this.framebuffer);
gl.framebufferTexture2D(
gl.FRAMEBUFFER, // The target is always a FRAMEBUFFER.
gl.COLOR_ATTACHMENT0, // We are providing the color buffer.
gl.TEXTURE_2D, // This is a 2D image texture.
texture, // The texture.
0); // 0, we aren't using MIPMAPs
gl.viewport(0, 0, width, height);
... // reading from output texture
gl.bindTexture(gl.TEXTURE_2D, texture);
gl.framebufferTexture2D(
gl.FRAMEBUFFER, gl.COLOR_ATTACHMENT0, gl.TEXTURE_2D, texture,
0);
gl.readPixels(0, 0, width, height, gl.FLOAT, gl.RED, buffer);
new answer
I'm just saying the same thing yet again (3rd time?)
Copied from below
WebGL is destination based. That means it's going to iterate over the pixels of the line/point/triangle it's drawing and for each point call the fragment shader and ask 'what value should I store here`?
It's destination based. It's going to draw each pixel exactly once. For that pixel it's going to ask "what color should I make this"
destination based loop
for (let i = start; i < end; ++i) {
fragmentShaderFunction(); // must set gl_FragColor
destinationTextureOrCanvas[i] = gl_FragColor;
You can see in the loop above there is no setting any random destination. There is no setting any part of destination twice. It's just going to run from start to end and exactly once for each pixel in the destination between start and end ask what color it should make that pixel.
How to do you set start and end? Again, to make it simple let's assume a 200x1 texture so we can ignore Y. It works like this
vertexShaderFunction(); // must set gl_Position
const start = clipspaceToArrayspaceViaViewport(viewport, gl_Position.x);
vertexShaderFunction(); // must set gl_Position
const end = clipspaceToArrayspaceViaViewport(viewport, gl_Position.x);
for (let i = start; i < end; ++i) {
fragmentShaderFunction(); // must set gl_FragColor
texture[i] = gl_FragColor;
}
see below for clipspaceToArrayspaceViaViewport
What is viewport? viewport is what you set when you called `gl.viewport(x, y, width, height)
So, set gl_Position.x to -1 and +1, viewport.x to 0 and viewport.width = 200 (the width of the texture) then start will be 0, end will be 200
set gl_Position.x to .25 and .75, viewport.x to 0 and viewport.width = 200 (the width of the texture). The start will be 125 and end will be 175
I honestly feel like this answer is leading you down the wrong path. It's not remotely this complicated. You don't have to understand any of this to use WebGL IMO.
The simple answer is
You set gl.viewport to the sub rectangle you want to affect in your destination (canvas or texture it doesn't matter)
You make a vertex shader that somehow sets gl_Position to clip space coordinates (they go from -1 to +1) across the texture
Those clip space coordinates get converted to the viewport space. It's basic math to map one range to another range but it's mostly not important. It's seems intuitive that -1 will draw to the viewport.x pixel and +1 will draw to the viewport.x + viewport.width - 1 pixel. That's what "maps from clip space to the viewport settings means".
It's most common for the viewport settings to be (x = 0, y = 0, width = width of destination texture or canvas, height = height of destination texture or canvas)
So that just leaves what you set gl_Position to. Those values are in clip space just like it explains in this article.
You can make it simple by doing if you want by converting from pixel space to clip space just like it explains in this article
zeroToOne = someValueInPixels / destinationDimensions;
zeroToTwo = zeroToOne * 2.0;
clipspace = zeroToTwo - 1.0;
gl_Position = clipspace;
If you continue the articles they'll also show adding a value (translation) and multiplying by a value (scale)
Using just those 2 things and a unit square (0 to 1) you can choose any rectangle on the screen. Want to effect 123 to 127. That's 5 units so scale = 5, translation = 123. Then apply the math above to convert from pixels to clips space and you'll get the rectangle you want.
If you continue further though those articles you'll eventually get the point where that math is done with matrices but you can do that math however you want. It's like asking "how do I compute the value 3". Well, 1 + 1 + 1, or 3 + 0, or 9 / 3, or 100 - 50 + 20 * 2 / 30, or (7^2 - 19) / 10, or ????
I can't tell you how to set gl_Position. I can only tell you make up whatever math you want and set it to *clip space* and then give an example of converting from pixels to clipspace (see above) as just one example of some possible math.
old answer
I get that this might not be clear I don't know how to help. WebGL draws lines, points, or triangles two a 2D array. That 2D array is either the canvas, a texture (as a framebuffer attachment) or a renderbuffer (as a framebuffer attachment).
The size of the area is defined by the size of the canvas, texture, renderbuffer.
You write a vertex shader. When you call gl.drawArrays(primitiveType, offset, count) you're telling WebGL to call your vertex shader count times. Assuming primitiveType is gl.TRIANGLES then for every 3 vertices generated by your vertex shader WebGL will draw a triangle. You specify that triangle by setting gl_Position in clip space.
Assuming gl_Position.w is 1, Clip space goes from -1 to +1 in X and Y across the destination canvas/texture/renderbuffer. (gl_Position.x and gl_Position.y are divided by gl_Position.w) which is not really important for your case.
To convert back to actually pixels your X and Y are converted based on the settings of gl.viewport. Let's just do X
pixelX = ((clipspace.x / clipspace.w) * .5 + .5) * viewport.width + viewport.x
WebGL is destination based. That means it's going to iterate over the pixels of the line/point/triangle it's drawing and for each point call the fragment shader and ask 'what value should I store here`?
Let's translate that to JavaScript in 1D. Let's assume you have an 1D array
const dst = new Array(100);
Let's make a function that takes a start and end and sets values between
function setRange(dst, start, end, value) {
for (let i = start; i < end; ++i) {
dst[i] = value;
}
}
You can fill the entire 100 element array with 123
const dst = new Array(100);
setRange(dst, 0, 99, 123);
To set the last half of the array to 456
const dst = new Array(100);
setRange(dst, 50, 99, 456);
Let's change that to use clip space like coordinates
function setClipspaceRange(dst, clipStart, clipEnd, value) {
const start = clipspaceToArrayspace(dst, clipStart);
const end = clipspaceToArrayspace(dst, clipEnd);
for (let i = start; i < end; ++i) {
dst[i] = value;
}
}
function clipspaceToArrayspace(array, clipspaceValue) {
// convert clipspace value (-1 to +1) to (0 to 1)
const zeroToOne = clipspaceValue * .5 + .5;
// convert zeroToOne value to array space
return Math.floor(zeroToOne * array.length);
}
This function now works just like the previous one except takes clip space values instead of array indices
// fill entire array with 123
const dst = new Array(100);
setClipspaceRange(dst, -1, +1, 123);
Set the last half of the array to 456
setClipspaceRange(dst, 0, +1, 456);
Now abstract one more time. Instead of using the array's length use a setting
// viewport looks like `{ x: number, width: number} `
function setClipspaceRangeViaViewport(dst, viewport, clipStart, clipEnd, value) {
const start = clipspaceToArrayspaceViaViewport(viewport, clipStart);
const end = clipspaceToArrayspaceViaViewport(viewport, clipEnd);
for (let i = start; i < end; ++i) {
dst[i] = value;
}
}
function clipspaceToArrayspaceViaViewport(viewport, clipspaceValue) {
// convert clipspace value (-1 to +1) to (0 to 1)
const zeroToOne = clipspaceValue * .5 + .5;
// convert zeroToOne value to array space
return Math.floor(zeroToOne * viewport.width) + viewport.x;
}
Now to fill the entire array with 123
const dst = new Array(100);
const viewport = { x: 0, width: 100; }
setClipspaceRangeViaViewport(dst, viewport, -1, 1, 123);
Set the last half of the array to 456 there are now 2 ways. Way one is just like the previous using 0 to +1
setClipspaceRangeViaViewport(dst, viewport, 0, 1, 456);
You can also set the viewport to start half way through the array
const halfViewport = { x: 50, width: 50; }
setClipspaceRangeViaViewport(dst, halfViewport, -1, +1, 456);
I don't know if that was helpful or not.
The only other thing to add is instead of value replace that with a function that gets called every iteration to supply value
function setClipspaceRangeViaViewport(dst, viewport, clipStart, clipEnd, fragmentShaderFunction) {
const start = clipspaceToArrayspaceViaViewport(viewport, clipStart);
const end = clipspaceToArrayspaceViaViewport(viewport, clipEnd);
for (let i = start; i < end; ++i) {
dst[i] = fragmentShaderFunction();
}
}
Note this is the exact same thing that is said in this article and clearified somewhat in this article.
I have what should be a simple exercise in OpenCV, but can't seem to get it working. I'm trying to determine the density of edges in a section of an image. This is the process I follow:
1. pull subimage from image
2. use Canny to find edges in subImage
3. threshold to create binary image
4. create histogram for binary image
5. get number of pixels in binary image that are "on" (255)
6. calculate "edge density" as numPixelsOn/totalPixels
I've checked the results of 1,2,and 3 above, and results seem ok. Steps 4 and 5 seem to be giving me trouble.
Here's my code for calculating the histogram:
int histSize = 256; // bin size
float range[] = { 0, 256} ;
const float* histRange = { range };
bool uniform = true;
bool accumulate = false;
Mat hist;
/// Compute the histograms:
calcHist( &gray, 1, 0, Mat(), hist, 1, &histSize, &histRange, uniform, accumulate );
This doesn't seem to be working. When I check hist after calling calcHist, it has no data (i.e. data == 0)... or maybe I don't understand what I'm looking at.
Now for accessing the "bins" in the histogram, I've tried a number of things. First I tried this:
uchar* p;
p = hist.ptr<uchar>(0);
double edgePixels = p[255];
I also tried to use:
cvQueryHistValue_1D(hist,255); // #include <opencv2/legacy/compat.hpp>
This wouldn't compile. Gave 2 errors: 'cv::Mat' does not have an overloaded member 'operator ->', and 'bins': is not a member of 'cv::Mat'
I guess I need some help on this.
There is an error in your 3rd param - channels, that should be an array so you should call it like this
int histSize = 256; // bin size
float range[] = { 0, 256} ;
const float* histRange = { range };
bool uniform = true;
bool accumulate = false;
Mat hist;
int channels[] = {0};
/// Compute the histograms:
calcHist( &gray, 1, channels, Mat(), hist, 1, &histSize, &histRange, uniform, accumulate );
You should also call:
hist.at<float>(0);
to get your value, OpenCV stores them as floats, this is the reason you're getting 0 when using uchar as uchar is smaller than float and the numbers stores as small enough to not fill the first bites.
This code is for 8 bit data to make gray-scale IplImage.
IplImage* img_gray_resize = NULL;
img_gray_resize = cvCreateImage(cvSize(320, 256), IPL_DEPTH_8U, 1);
DWORD dwCount;
LVDS_SetDataMode(0); // o for 8 bit mode and 1 for 16 bit mode
dwCount = (LONG)320 * (LONG)256;
unsigned char* m_pImage = NULL;
m_pImage = new unsigned char[320 * 256];
for (int i=0; i<320 * 256; i++) m_pImage[i] = NULL;
LVDS_GetFrame(&dwCount, m_pImage);
int width = 320;
int height = 256;
int nn = 0;
int ii = 0;
for (int y=0; y<height; y++)
{
for (int x=0; x<width; x++)
{
ii = y * width + x;
if(nn < (height*width))
img_gray_resize->imageData[ii] = m_pImage[nn++];
}
}
delete [] m_pImage;
I need to display 16 bit gray-scale image. If I display 8 bit data, some information is missing from the image. However, LVDS_SetDataMode() can provide both types of data. I am using a library for frame grabber device. Please help me.
16 bit images should be stored in IPL_DEPTH_16U (or CV_16U) mode. This is the correct memory layout.
However, displaying them depends on your display hardware.
Most regular display APIs, e.g. OpenCV's highgui, can only display 8-bit images.
To actually display the image, you will have to convert your image to 8-bits for display.
You will need to decide how to do this. There are many ways to do this, depending on your application and complexity. Some options are:
Show MSB = right-shift the image by 8 pixels.
Show LSB = saturate anything above 255.
In fact, right-shift by any value between 0-8 bits, combined with a cv::saturate_cast to avoid value wrap-around.
HDR->LDR = Apply dynamic range compression algorithms.
as I know,only 8bit data can be displayed,you need to find the best way to convert the 16bit to 8bit to minimize the information you lose. Histogram equalization can be applyed to do this.
Finally, I have solved the problem by following way:
dwCount = (LONG)320 * (LONG)256 * 2;
LVDS_SetDataMode(1);
img_gray_resize->imageData[ii] = m_pImage[nn++] >> 6;
Just shift bits to right (2, 3, 4, 5, 6, ...), where you get good result, use that value.
I'm learning cuda texture memory. Now, I got a opencv Iplimage, and I get its imagedata. Then I bind a texture to this uchar array, like below:
Iplimage *image = cvCreateImage(cvSize(width, height), IPL_DEPTH_8U, 3);
unsigned char* imageDataArray = (unsigned char*)image->imagedata;
texture<unsigned char,2,cudaReadModeElementType> tex;
cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc(8, 8, 8, 0,
cudaChannelFormatKindUnsigned);
cudaArray *cuArray = NULL;
CudaSafeCall(cudaMallocArray(&cuArray,&channelDesc,width,height));
cudaMemcpy2DToArray(cuArray,0,0,imageDataArray,image->widthstep,
width * sizeof(unsigned char), height, cudaMemcpyHostToDevice);
cudaBindTextureToArray(texC1_cf,cuArray_currentFrame, channelDesc);
Now I lanch my kernel, and I want to access each pixel, every channel of that image. This is where I get confused.
I use this code to get the pixel coordinate (X,Y):
int X = (blockIdx.x*blockDim.x+threadIdx.x);
int Y = (blockIdx.y*blockDim.y+threadIdx.y);
And how can I access each channel of this (X,Y)? what's the code below return?
tex2D(tex, X, Y);
Besides this, Can you tell me how texture memory using texture to access an array, and how this transform looks like?
To bind a 3 channel OpenCV image to cudaArray texture, you have to create a cudaArray of width equal to image->width * image->nChannels, because the channels are stored interleaved by OpenCV.
cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc<unsigned char>();
cudaArray *cuArray = NULL;
CudaSafeCall(cudaMallocArray(&cuArray,&channelDesc,width * image->nChannels,height));
cudaMemcpy2DToArray(cuArray,0,0,imageDataArray,image->widthstep, width * image->nChannels * sizeof(unsigned char), height, cudaMemcpyHostToDevice);
cudaBindTextureToArray(texC1_cf,cuArray_currentFrame, channelDesc);
Now, to access each channel separately in the kernel, you just have to multiply the x index with number of channels and add the offset of desired channel like this:
unsigned char blue = tex2D(tex, (3 * X) , Y);
unsigned char green = tex2D(tex, (3 * X) + 1, Y);
unsigned char red = tex2D(tex, (3 * X) + 2, Y);
First one is blue because OpenCV stores images with channel sequence BGR.
As for the error you get when you try to access texture<uchar3,..> using tex2D; CUDA only supports creating 2D textures of 1,2 and 4 element vector types. Unfortunately, ONLY 3 is not supported which is very good for binding RGB images and is a really desirable feature.
I need to convert an 8-bit IplImage to a 32-bits IplImage. Using documentation from all over the web I've tried the following things:
// general code
img2 = cvCreateImage(cvSize(img->width, img->height), 32, 3);
int height = img->height;
int width = img->width;
int channels = img->nChannels;
int step1 = img->widthStep;
int step2 = img2->widthStep;
int depth1 = img->depth;
int depth2 = img2->depth;
uchar *data1 = (uchar *)img->imageData;
uchar *data2 = (uchar *)img2->imageData;
for(h=0;h<height;h++) for(w=0;w<width;w++) for(c=0;c<channels;c++) {
// attempt code...
}
// attempt one
// result: white image, two red spots which appear in the original image too.
// this is the closest result, what's going wrong?!
// see: http://files.dazjorz.com/cache/conversion.png
((float*)data2+h*step2+w*channels+c)[0] = data1[h*step1+w*channels+c];
// attempt two
// when I change float to unsigned long in both previous examples, I get a black screen.
// attempt three
// result: seemingly random data to the top of the screen.
data2[h*step2+w*channels*3+c] = data1[h*step1+w*channels+c];
data2[h*step2+w*channels*3+c+1] = 0x00;
data2[h*step2+w*channels*3+c+2] = 0x00;
// and then some other things. Nothing did what I wanted. I couldn't get an output
// image which looked the same as the input image.
As you see I don't really know what I'm doing. I'd love to find out, but I'd love it more if I could get this done correctly.
Thanks for any help I get!
The function you are looking for is cvConvertScale(). It automagically does any type conversion for you. You just have to specify that you want to scale by a factor of 1/255 (which maps the range [0...255] to [0...1]).
Example:
IplImage *im8 = cvLoadImage(argv[1]);
IplImage *im32 = cvCreateImage(cvSize(im8->width, im8->height), 32, 3);
cvConvertScale(im8, im32, 1/255.);
Note the dot in 1/255. - to force a double division. Without it you get a scale of 0.
Perhaps this link can help you?
Edit In response to the second edit of the OP and the comment
Have you tried
float value = 0.5
instead of
float value = 0x0000001;
I thought the range for a float color value goes from 0.0 to 1.0, where 1.0 is white.
Floating point colors go from 0.0 to 1.0, and uchars go from 0 to 255. The following code fixes it:
// h is height, w is width, c is current channel (0 to 2)
int b = ((uchar *)(img->imageData + h*img->widthStep))[w*img->nChannels + c];
((float *)(img2->imageData + h*img2->widthStep))[w*img2->nChannels + c] = ((float)b) / 255.0;
Many, many thanks to Stefan Schmidt for helping me fix this!
If you do not put the dot (.), some compilers will understand is as an int division, giving you a int result (zero in this case).
You can create an IplImage wrapper using boost::shared_ptr and template-metaprogramming. I have done that, and I get automatic garbage collection, together with automatic image conversions from one depth to another, or from one-channel to multi-channel images.
I have called the API blImageAPI and it can be found here:
http://www.barbato.us/2010/10/14/image-data-structure-based-shared_ptr-iplimage/
It is very fast, and make code very readable, (good for maintaining algorithms)
It is also can be used instead of IplImage in opencv algorithms without changing anything.
Good luck and have fun writing algorithms!!!
IplImage *img8,*img32;
img8 =cvLoadImage("a.jpg",1);
cvNamedWindow("Convert",1);
img32 = cvCreateImage(cvGetSize(img8),IPL_DEPTH_32F,3);
cvConvertScale(img8,img32,1.0/255.0,0.0);
//For Confirmation Check the pixel values (between 0 - 1)
for(int row = 0; row < img32->height; row++ ){
float* pt = (float*) (img32->imageData + row * img32->widthStep);
for ( int col = 0; col < width; col++ )
printf("\n %3.3f , %3.3f , %3.3f ",pt[3*col],pt[3*col+1],pt[3*col+2]);
}
cvShowImage("Convert",img32);
cvWaitKey(0);
cvReleaseImage(&img8);
cvReleaseImage(&img32);
cvDestroyWindow("Convert");