Display data from pnglib as an ximage - xlib

I need to import a PNG and display it on screen in a Motif application. For reasons best known to myself, I don't want to use any more libraries than I need to, and I'd like to stick with just Motif and pnglib.
I've been battling with this for a couple of days now, and I'd like to put aside my pride and ask for some help. This screenshot shows the problem:
https://s3.amazonaws.com/gtrebol264929/pnglib_fail.png
The window on the right shows what the image should look like, the window on the left is my Motif application showing what it looks like in my app. Clearly I've got the image data OK, as the basic concept of the picture can be seen. But also clearly I've messed up how I get the pixel data from pnglib into an XImage. Below is my code:
char * xdata = malloc(width * height * (channels + 1));
memset(xdata,100,width * height * channels);
int colc = 0;
int bytec = 0;
while (colc < width) {
int rowc = 0;
while(rowc < height) {
png_byte * row = png.row_pointers[rowc];
memcpy(&xdata[bytec],&row[colc],1);
bytec += 4;
rowc += 1;
}
colc += 1;
}
XImage * img = XCreateImage(display, CopyFromParent, depth * channels, ZPixmap, 0, xdata, width, height, 32, bytes_per_line);
printf("PNG %ix%i (depth: %i x %i) img: %p\n",width,height,depth,channels,img);
XPutImage (display, win, gc, img, 0, 0, 0, 0, width, height); // 0, 0, 0, 0 are src x,y and dst x,y
png.row_pointers is the pixel data from pnglib.
I'm pretty sure I've just misunderstood how the pixel data is stored, but I can't quite work out what I've done wrong. Any help is very much appreciated.
All the best
Garry

Related

Very unexpected behavior of C++ win32 BitBlt

I noticed when I try to run BitBlt, the resulting data buffer is unexpected in two ways:
It is flipped along the y axis (the origin seems to be bottom left instead of top left)
In each RGBA grouping, the R and B values seem to be switched.
For the first issue, I noticed it when testing with my command prompt; if my command prompt was in the upper left portion of the screen, it would only say it was black when my cursor was in the lower left portion. I had to fix the inversion of the y axis by changing int offset = (y * monitor_width + x) * 4; to int offset = ((monitor_height - 1 - y) * monitor_width + x) * 4; this fixed the pixel location issue because it was showing black where I expected black.
However, the colors were still strong. I tested by trying to get the color of known pixels. I noticed every blue pixel had a very high R value and every red pixel had a very high blue value. That's when I compared with an existing tool I had and found out that the red and blue values seem to be switched in every pixel. At first I thought it was backwards or a byte alignment issue, but I also verified in a clustering of pixels that aren't uniform to make sure it's picking the right position of pixel, and it did perfectly well, just with the colors switched.
Full simplified code below (originally my tool was getting my cursor position and printing the pixel color via hotkey press; this is a simplified version that gets one specific point).
BYTE* my_pixel_data;
HDC hScreenDC = GetDC(GetDesktopWindow());
int BitsPerPixel = GetDeviceCaps(hScreenDC, BITSPIXEL);
HDC hMemoryDC = CreateCompatibleDC(hScreenDC);
int monitor_width = GetSystemMetrics(SM_CXSCREEN);
int monitor_height = GetSystemMetrics(SM_CYSCREEN);
std::cout << std::format("monitor width height: {}, {}\n", monitor_width, monitor_height);
BITMAPINFO info;
info.bmiHeader.biSize = sizeof(BITMAPINFOHEADER);
info.bmiHeader.biWidth = monitor_width; // client_width;
info.bmiHeader.biHeight = monitor_height; // client_height;
info.bmiHeader.biPlanes = 1;
info.bmiHeader.biBitCount = BitsPerPixel;
info.bmiHeader.biCompression = BI_RGB;
HBITMAP hbitmap = CreateDIBSection(hMemoryDC, &info, DIB_RGB_COLORS, (void**)&my_pixel_data, 0, 0);
SelectObject(hMemoryDC, hbitmap);
BitBlt(hMemoryDC, 0, 0, monitor_width, monitor_height, hScreenDC, 0, 0, SRCCOPY);
int x = 12, y = 12;
int offset = ((monitor_height - 1 - y) * monitor_width + x) * 4;
std::cout << std::format("debug: ({}, {}): ({}, {}, {})\n", x, y, (int)my_pixel_data[offset], (int)my_pixel_data[offset + 1], (int)my_pixel_data[offset + 2], (int)my_pixel_data[offset + 3]);
system("pause");
The output of this will be debug: (12, 12): (199, 76, 133) even though another program has verified the colors are actually (133, 76, 199).
I can easily fix this in my code by flipping the y axis and switching each R and B value and the program will work perfectly well. However, I am just baffled by how this happened and whether there's a more elegant fix.
I can answer the RGB (and it looks like Hans answered the inverted Y axis in a comment). Remember that RGB is stored 0xAARRGGBB, so in that 32 bit value BB is byte 0, GG is byte 1, and RR is byte 2 (alpha is byte 3 if you use it), so when you index in at +0, +1 and +2 you're actually getting the values correctly. When we say RGB we're saying the colors in opposite order of how they're stored in memory.

Processing: Efficiently create uniform grid

I'm trying to create a grid of an image (in the way one would tile a background with). Here's what I've been using:
PImage bgtile;
PGraphics bg;
int tilesize = 50;
void setup() {
int t = millis();
fullScreen(P2D);
background(0);
bgtile = loadImage("bgtile.png");
int bgw = ceil( ((float) width) / tilesize) + 1;
int bgh = ceil( ((float) height) / tilesize) + 1;
bg = createGraphics(bgw*tilesize,bgh*tilesize);
bg.beginDraw();
for(int i = 0; i < bgw; i++){
for(int j = 0; j < bgh; j++){
bg.image(bgtile, i*tilesize, j*tilesize, tilesize, tilesize);
}
}
bg.endDraw();
print(millis() - t);
}
The timing code says that this takes about a quarter of a second, but by my count there's a full second once the window opens before anything shows up on screen (which should happen as soon as draw is first run). Is there a faster way to get this same effect? (I want to avoid rendering bgtile hundreds of times in the draw loop for obvious reasons)
One way could be to make use of the GPU and let OpenGL repeat a texture for you.
Processing makes it fairly easy to repeat a texture via textureWrap(REPEAT)
Instead of drawing an image you'd make your own quad shape and instead of calling vertex(x, y) for example, you'd call vertex(x, y, u, v); passing texture coordinates (more low level info on the OpenGL link above). The simple idea is x,y would control the geometry on screen and u,v would control how the texture is applied to the geometry.
Another thing you can control is textureMode() which allows you control how you specify the texture coordinates (U, V):
IMAGE mode is the default: you use pixel coordinates (based on the dimensions of the texture)
NORMAL mode uses values between 0.0 and 1.0 (also known as normalised values) where 1.0 means the maximum the texture can go (e.g. image width for U or image height for V) and you don't need to worry about knowing the texture image dimensions
Here's a basic example based on the textureMode() example above:
PImage img;
void setup() {
fullScreen(P2D);
noStroke();
img = loadImage("https://processing.org/examples/moonwalk.jpg");
// texture mode can be IMAGE (pixel dimensions) or NORMAL (0.0 to 1.0)
// normal means 1.0 is full width (for U) or height (for V) without having to know the image resolution
textureMode(NORMAL);
// this is what will make handle tiling for you
textureWrap(REPEAT);
}
void draw() {
// drag mouse on X axis to change tiling
int tileRepeats = (int)map(constrain(mouseX,0,width), 0, width, 1, 100);
// draw a textured quad
beginShape(QUAD);
// set the texture
texture(img);
// x , y , U , V
vertex(0 , 0 , 0 , 0);
vertex(width, 0 , tileRepeats, 0);
vertex(width, height, tileRepeats, tileRepeats);
vertex(0 , height, 0 , tileRepeats);
endShape();
text((int)frameRate+"fps",15,15);
}
Drag the mouse on the Y axis to control the number of repetitions.
In this simple example both vertex coordinates and texture coordinates are going clockwise (top left, top right, bottom right, bottom left order).
There are probably other ways to achieve the same result: using a PShader comes to mind.
Your approach caching the tiles in setup is ok.
Even flattening your nested loop into a single loop at best may only shave a few milliseconds off, but nothing substantial.
If you tried to cache my snippet above it would make a minimal difference.
In this particular case, because of the back and forth between Java/OpenGL (via JOGL), as far as I can tell using VisualVM, it looks like there's not a lot of room for improvement since simply swapping buffers takes so long (e.g. bg.image()):
An easy way to do this would be to use processing's built in get(); which saves a PImage of the coordinates you pass, for example: PImage pic = get(0, 0, width, height); will capture a "screenshot" of your entire window. So, you can create the image like you already are, and then take a screenshot and display that screenshot.
PImage bgtile;
PGraphics bg;
PImage screenGrab;
int tilesize = 50;
void setup() {
fullScreen(P2D);
background(0);
bgtile = loadImage("bgtile.png");
int bgw = ceil(((float) width) / tilesize) + 1;
int bgh = ceil(((float) height) / tilesize) + 1;
bg = createGraphics(bgw * tilesize, bgh * tilesize);
bg.beginDraw();
for (int i = 0; i < bgw; i++) {
for (int j = 0; j < bgh; j++) {
bg.image(bgtile, i * tilesize, j * tilesize, tilesize, tilesize);
}
}
bg.endDraw();
screenGrab = get(0, 0, width, height);
}
void draw() {
image(screenGrab, 0, 0);
}
This will still take a little bit to generate the image, but once it does, there is no need to use the for loops again unless you change the tilesize.
#George Profenza's answer looks more efficient than my solution, but mine may take a little less modification to the code you already have.

Direct Access to CreateDIBitmap Bits

[The final fix, which works unconditionally: use SetDIBitsToDevice, not BitBlt, to copy out the post-text-draw image data. With this change, all occurrences of the problem are gone.]
I fixed the problem I'm having, but for the life of me I can't figure out why it occurred.
Create a bitmap with CreateDIBitmap. Get a pointer to the bitmap bits.
Select the bitmap into a memory DC.
Background fill the bitmap by directly writing the bitmap memory.
TextOut.
No text displays.
What fixed the problem: change item 3. from direct fill to a call to FillRect. All is well, it works perfectly.
This is under Windows 10 but from what little I could find on the web, it spans all versions of Windows. NO operations work on the bitmap - even calling FillRect - after the manual write. No savvy, Kimosabe. Elsewhere in the app, I even build gradient fills by directly writing to that bitmap memory and there is no problem. But once TextOut is called after the manual fill, the bitmap is locked (effectively) and no further functions work on it - nor do any return an error.
I'm using a font with a 90 degree escapement. Have not tried it with a "normal" font, 0 degree escapement. DrawTextEx with DT_CALCRECT specifically states it only works on 0 degree escapement fonts so I had to use TextOut for this reason.
Very bizarre.
No, there were no stupid mistakes like using the same text color as the background color. I've spent too long on this for that. One option people have available is that the endless energy that would normally be spent destroying the question and/or the person who asked it could instead be used to write a few lines of code and try it for yourself.
Here's a function to make a bitmap. Don't pass a plain colour, pass a gradient fill, say going from white to pinkish.
Does it display correctly? If so, does the TextOut call on top of that work?
static HBITMAP MakeBitmap(unsigned char *rgba, int width, int height, VOID **buff)
{
VOID *pvBits; // pointer to DIB section
HBITMAP answer;
BITMAPINFO bmi;
HDC hdc;
int x, y;
int red, green, blue, alpha;
// setup bitmap info
bmi.bmiHeader.biSize = sizeof(BITMAPINFOHEADER);
bmi.bmiHeader.biWidth = width;
bmi.bmiHeader.biHeight = height;
bmi.bmiHeader.biPlanes = 1;
bmi.bmiHeader.biBitCount = 32; // four 8-bit components
bmi.bmiHeader.biCompression = BI_RGB;
bmi.bmiHeader.biSizeImage = width * height * 4;
hdc = CreateCompatibleDC(GetDC(0));
answer = CreateDIBSection(hdc, &bmi, DIB_RGB_COLORS, &pvBits, NULL, 0x0);
for (y = 0; y < height; y++)
{
for (x = 0; x < width; x++)
{
red = rgba[(y*width + x) * 4];
green = rgba[(y*width + x) * 4 + 1];
blue = rgba[(y*width + x) * 4 + 2];
alpha = rgba[(y*width + x) * 4 + 3];
red = (red * alpha) >> 8;
green = (green * alpha) >> 8;
blue = (blue * alpha) >> 8;
((UINT32 *)pvBits)[(height - y - 1) * width + x] = (alpha << 24) | (red << 16) | (green << 8) | blue;
}
}
DeleteDC(hdc);
*buff = pvBits;
return answer;
}

CreateDIBSection to get uint8_t[] of portion of screen

I am trying to get a uint8_t[] of a portion of the screen. The xy coordinates of the top left is 2,3 and of bottom right is 17,18.
This is the top 30x30 pixels of my screen with regular screenshot and photoshop crop:
And this is what 2,3 to 17,18 screenshot should look like via regular screenshot and Photoshop crop:
This is what I am getting as a result of my code:
My code is in js-ctypes but there is no ctypes errors. This is a winapi thing. So I didn't tag this with ctypes as they would be confused. This is the simplified code I am using, the error checks etc have been removed:
c1 = {x:2, y:3} // top left corner
c2 = {x:17, y:18} // bottom right corner
CreateDC('DISPLAY', null, null, null);
nBPP = GetDeviceCaps(hdcScreen, BITSPIXEL);
w = c2.x - c1.x; // width = 15
h = c2.y - c1.y; // height = 15
hdcMemoryDC = CreateCompatibleDC(hdcScreen);
bmi = BITMAPINFO();
bmi.bmiHeader.biSize = BITMAPINFOHEADER.size;
bmi.bmiHeader.biWidth = w;
bmi.bmiHeader.biHeight = -1 * h;
bmiHeader.biPlanes = 1;
bmi.bmiHeader.biBitCount = nBPP; // nBPP is 32
bmi.bmiHeader.biCompression = BI_RGB;
hbmp = CreateDIBSection(hdcScreen, &bmi, DIB_RGB_COLORS, (void**)&pixelBuffer, null, 0);
SelectObject(hdcMemoryDC, hbmp);
BitBlt(hdcMemoryDC, 0, 0, w, h, hdcScreen, c1.x, c1.y, SRCCOPY);
Why portion of screen bits come out wrong? If I do a full screen shot it works fine.

QT QImage pixel manipulation

I am building a QT GUI application and use QImage for opening images.
My problem is that I can't figure out how to use QImage's bit() and scanline()
methods to get access at per pixel level.
I've seen this post Qt QImage pixel manipulation problems
but this is only for the first pixel of each row. Is this correct or I got it all wrong?
thanks in advance
The scanlines correspond to the the height of image, the columns correspond to the width of the image.
According to the docs, the prototype looks like uchar* QImage::scanline(int i), or a similar const version.
But, as a commenter pointed out, because the data is dependent on the machine architecture and image, you should NOT use the uchar * directly. Instead, use something like the following:
QRgb *rowData = (QRgb*)img.scanLine(row);
QRgb pixelData = rowData[col];
int red = qRed(pixelData);
It may not be immediately obvious from Kaleb's post, but the following works for setting a pixel on a Format_RGB32 image.
// Get the line we want
QRgb *line = (QRgb *)image->scanLine(row_index);
// Go to the pixel we want
line += col_index;
// Actually set the pixel
*line = qRgb(qRed(color), qGreen(color), qBlue(color));
The answer did not work for me. It looks like, the data is not 32bit aligned on my system.
To get the correct data, on my system i had to do this:
for(uint32_t Y = 0; Y < mHeight; ++Y)
{
uint8_t* pPixel = Image.scanLine(Y);
for(uint32_t X = 0; X < mWidth; ++X)
{
const int Blue = *pPixel++;
const int Green = *pPixel++;
const int Red = *pPixel++;
uint8_t GrayscalePixel = (0.21f * Red) + (0.72f * Green) + (0.07 * Blue);
}
}

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