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Optimization of conversion from opencv mat/Array to to OnnxRuntime Tensor?

I am using the ONNXRuntime to inference a UNet model and as a part of preprocessing I have to convert an EMGU OpenCV matrix to OnnxRuntime.Tensor.
I achieved it using two nested for loops which is unfortunately quite slow:
var data = new DenseTensor<float>(new[] { 1, 3, WIDTH, HEIGHT});
for (int y = 0; y < HEIGHT; y++)
{
for (int x = 0; x < WIDTH; x++)
{
data[0, 0, x, y] = image.GetValue(2, y, x)/255.0;
data[0, 1, x, y] = image.GetValue(1, y, x)/255.0;
data[0, 2, x, y] = image.GetValue(0, y, x)/255.0;
}
}
Then I found out that there exists a method which converts Array to DenseTensor. I wanted to use this method as follows:
var imgToPredictFloat = new Mat(image.Height, image.Width, DepthType.Cv32F, 3);
image.ConvertTo(imgToPredictFloat, DepthType.Cv32F, 1/255.0);
CvInvoke.CvtColor(imgToPredictFloat, imgToPredictFloat, ColorConversion.Bgra2Rgb);
var data = image.GetData().ToTensor<float>;
var reshaped = data.Reshape(new int[] { 1, 3, WIDTH, HEIGHT});
This would greatly improve the performance however the layout of the output tensor is not correct (the same as from the for loop) and the model obviously won't work. Any suggestions how to reshape the array to the correct layout?
In the code is also performed converting int 0-255 to float 0-1 and BGR layout to RGB layout.

This is how I have used cv::Mat with ONNX Runtime ( C++ ) :
const wchar_t* model_path = L"C:/data/DNN/ONNX/ResNet/resnet152v2/resnet152-v2-7.onnx";
printf("Using Onnxruntime C++ API\n");
Ort::Session session(env, model_path, session_options);
//*************************************************************************
// print model input layer (node names, types, shape etc.)
Ort::AllocatorWithDefaultOptions allocator;
size_t num_output_nodes = session.GetOutputCount();
std::vector<char*> outputNames;
for (size_t i = 0; i < num_output_nodes; ++i)
{
char* name = session.GetOutputName(i, allocator);
std::cout << "output: " << name << std::endl;
outputNames.push_back(name);
}
// print number of model input nodes
size_t num_input_nodes = session.GetInputCount();
std::vector<const char*> input_node_names(num_input_nodes);
std::vector<int64_t> input_node_dims; // simplify... this model has only 1 input node {1, 3, 224, 224}.
// Otherwise need vector<vector<>>
printf("Number of inputs = %zu\n", num_input_nodes);
// iterate over all input nodes
for (int i = 0; i < num_input_nodes; i++) {
// print input node names
char* input_name = session.GetInputName(i, allocator);
printf("Input %d : name=%s\n", i, input_name);
input_node_names[i] = input_name;
// print input node types
Ort::TypeInfo type_info = session.GetInputTypeInfo(i);
auto tensor_info = type_info.GetTensorTypeAndShapeInfo();
ONNXTensorElementDataType type = tensor_info.GetElementType();
printf("Input %d : type=%d\n", i, type);
// print input shapes/dims
input_node_dims = tensor_info.GetShape();
printf("Input %d : num_dims=%zu\n", i, input_node_dims.size());
for (int j = 0; j < input_node_dims.size(); j++)
printf("Input %d : dim %d=%jd\n", i, j, input_node_dims[j]);
}
cv::Size dnnInputSize;
cv::Scalar mean;
cv::Scalar std;
bool rgb = true;
//cv::Mat inputImage = cv::imread("C:/TestImages/kitten_01.jpg");
cv::Mat inputImage = cv::imread("C:/TestImages/slug_01.jpg");
rgb = true;
dnnInputSize = cv::Size(224, 224);
mean[0] = 0.485;
mean[1] = 0.456;
mean[2] = 0.406;
std[0] = 0.229;
std[1] = 0.224;
std[2] = 0.225;
cv::Mat blob;
// ONNX: (N x 3 x H x W)
cv::dnn::blobFromImage(inputImage, blob, 1.0 / 255.0, dnnInputSize, mean, rgb, false);
size_t input_tensor_size = blob.total();
std::vector<float> input_tensor_values(input_tensor_size);
for (size_t i = 0; i < input_tensor_size; ++i)
{
input_tensor_values[i] = blob.at<float>(i);
}
std::vector<const char*> output_node_names = { outputNames.front() };
// create input tensor object from data values
auto memory_info = Ort::MemoryInfo::CreateCpu(OrtArenaAllocator, OrtMemTypeDefault);
Ort::Value input_tensor = Ort::Value::CreateTensor<float>(memory_info, input_tensor_values.data(), input_tensor_size, input_node_dims.data(), 4);
assert(input_tensor.IsTensor());
// score model & input tensor, get back output tensor
auto output_tensors = session.Run(Ort::RunOptions{ nullptr }, input_node_names.data(), &input_tensor, 1, output_node_names.data(), 1);
assert(output_tensors.size() == 1 && output_tensors.front().IsTensor());
// Get pointer to output tensor float values
float* floatarr = output_tensors.front().GetTensorMutableData<float>();
assert(abs(floatarr[0] - 0.000045) < 1e-6);
cv::Mat1f result = cv::Mat1f(1000, 1, floatarr);
cv::Point classIdPoint;
double confidence = 0;
minMaxLoc(result, 0, &confidence, 0, &classIdPoint);
int classId = classIdPoint.y;
std::cout << "confidence: " << confidence << std::endl;
std::cout << "class: " << classId << std::endl;
The actual conversion part that you need is imho (adjust size and mean/std according to your network):
cv::Mat inputImage = cv::imread("C:/TestImages/slug_01.jpg");
rgb = true;
dnnInputSize = cv::Size(224, 224);
mean[0] = 0.485;
mean[1] = 0.456;
mean[2] = 0.406;
std[0] = 0.229;
std[1] = 0.224;
std[2] = 0.225;
cv::Mat blob;
// ONNX: (N x 3 x H x W)
cv::dnn::blobFromImage(inputImage, blob, 1.0 / 255.0, dnnInputSize, mean, rgb, false);

C# and Kinect v2: Get RGB values that fit to depth-pixel

I played a bit around with the Kinect v2 and C# and tried to get a 512x424 pixel-sized image array that contains depth data aswell as the regarding color information (RGBA).
Therefore I used the MultiSourceFrameReader class to receive a MultiSourceFrame from which I got the ColorFrame and DepthFrame. With the methods ColorFrame.CopyConvertedFrameDataToArray() and DepthFrame.CopyFrameDataToArray() I received the arrays that hold color and depth information:
// Contains 4*1920*1080 entries of color-info: BGRA|BGRA|BGRA..
byte[] cFrameData = new byte[4 * cWidth * cHeight];
cFrame.CopyConvertedFrameDataToArray(cFrameData, ColorImageFormat.Bgra);
// Has 512*424 entries with depth information
ushort[] dFrameData = new ushort[dWidth* dHeight];
dFrame.CopyFrameDataToArray(dFrameData);
Now I would have to map the color-quadruples that live within the ColorFrame-data-array cFrameData to each of the entries of the DepthFrame-data-array dFrameData but that's where I'm stuck. Output should be an array that is 4 times (RGBA/BGRA) the size of the dFrameData array and contains the color information to each pixel of the depth-frame:
// Create the array that contains the color information for every depth-pixel
byte[] dColors = new byte[4 * dFrameData.Length];
for (int i = 0, j = 0; i < cFrameData.Length; ++i)
{
// The mapped color index. ---> I'm stuck here:
int colIx = ?;
dColors[j] = cFrameData[colIx]; // B
dColors[j + 1] = cFrameData[colIx + 1]; // G
dColors[j + 2] = cFrameData[colIx + 2]; // R
dColors[j + 3] = cFrameData[colIx + 3]; // A
j += 4;
}
Does anyone have any suggestions?
I also took a look at the Kinect-SDK's CoordinateMappingBasics example but they did it vice versa for the 1920x1080 pixel-sized image which I already got to work.
Edit
I recognized that I should be able to get the mapped color information by using the ColorSpacePoint-struct which contains the X and Y coordinates to the specific color pixel. Therefore I set up the points like..
// Lookup table for color-point information
ColorSpacePoint[] cSpacePoints = new ColorSpacePoint[dWidth * dHeight];
this.kinectSensor.CoordinateMapper.MapDepthFrameToColorSpace(dFrameData, cSpacePoints);
.. and tried to access the color information like ..
int x = (int)(cSpacePoints[i].X + 0.5f);
int y = (int)(cSpacePoints[i].Y + 0.5f);
int ix = x * cWidth + y;
byte r = cFrameData[ix + 2];
byte g = cFrameData[ix + 1];
byte b = cFrameData[ix];
byte a = cFrameData[ix + 3];
.. but I'm still getting the wrong colors. Mostly white ones.

Well, I figured it out by myself. The error was trivial. As the array is not a pixel-array where one entry contains RGBA information but a byte array where each entry represents either R, G, B or A I had to multiply the index by the bytes-per-pixel value which in this case is 4. So the solution looks like:
int ix = (x * cWidth + y) * 4;
byte r = cFrameData[ix + 2];
byte g = cFrameData[ix + 1];
byte b = cFrameData[ix];
byte a = cFrameData[ix + 3];

fast way to convert integer array to byte array (11 bit)

I have integer array and I need to convert it to byte array
but I need to take (only and just only) first 11 bit of each element of the هinteger array
and then convert it to a byte array
I tried this code
// ***********convert integer values to byte values
//***********to avoid the left zero padding on the byte array
// *********** first step : convert to binary string
// ***********second step : convert binary string to byte array
// *********** first step
string ByteString = Convert.ToString(IntArray[0], 2).PadLeft(11,'0');
for (int i = 1; i < IntArray.Length; i++)
ByteString = ByteString + Convert.ToString(IntArray[i], 2).PadLeft(11, '0');
// ***********second step
int numOfBytes = ByteString.Length / 8;
byte[] bytes = new byte[numOfBytes];
for (int i = 0; i < numOfBytes; ++i)
{
bytes[i] = Convert.ToByte(ByteString.Substring(8 * i, 8), 2);
}
But it takes too long time (if the file size large , the code takes more than 1 minute)
I need a very very fast code (very few milliseconds only )
can any one help me ?

Basically, you're going to be doing a lot of shifting and masking. The exact nature of that depends on the layout you want. If we assume that we pack little-endian from each int, appending on the left, so two 11-bit integers with positions:
abcdefghijk lmnopqrstuv
become the 8-bit chunks:
defghijk rstuvabc 00lmnopq
(i.e. take the lowest 8 bits of the first integer, which leaves 3 left over, so pack those into the low 3 bits of the next byte, then take the lowest 5 bits of the second integer, then finally the remaining 6 bits, padding with zero), then something like this should work:
using System;
using System.Linq;
static class Program
{
static string AsBinary(int val) => Convert.ToString(val, 2).PadLeft(11, '0');
static string AsBinary(byte val) => Convert.ToString(val, 2).PadLeft(8, '0');
static void Main()
{
int[] source = new int[1432];
var rand = new Random(123456);
for (int i = 0; i < source.Length; i++)
source[i] = rand.Next(0, 2047); // 11 bits
// Console.WriteLine(string.Join(" ", source.Take(5).Select(AsBinary)));
var raw = Encode(source);
// Console.WriteLine(string.Join(" ", raw.Take(6).Select(AsBinary)));
var clone = Decode(raw);
// now prove that it worked OK
if (source.Length != clone.Length)
{
Console.WriteLine($"Length: {source.Length} vs {clone.Length}");
}
else
{
int failCount = 0;
for (int i = 0; i < source.Length; i++)
{
if (source[i] != clone[i] && failCount++ == 0)
{
Console.WriteLine($"{i}: {source[i]} vs {clone[i]}");
}
}
Console.WriteLine($"Errors: {failCount}");
}
}
static byte[] Encode(int[] source)
{
long bits = source.Length * 11;
int len = (int)(bits / 8);
if ((bits % 8) != 0) len++;
byte[] arr = new byte[len];
int bitOffset = 0, index = 0;
for (int i = 0; i < source.Length; i++)
{
// note: this encodes little-endian
int val = source[i] & 2047;
int bitsLeft = 11;
if(bitOffset != 0)
{
val = val << bitOffset;
arr[index++] |= (byte)val;
bitsLeft -= (8 - bitOffset);
val >>= 8;
}
if(bitsLeft >= 8)
{
arr[index++] = (byte)val;
bitsLeft -= 8;
val >>= 8;
}
if(bitsLeft != 0)
{
arr[index] = (byte)val;
}
bitOffset = bitsLeft;
}
return arr;
}
private static int[] Decode(byte[] source)
{
int bits = source.Length * 8;
int len = (int)(bits / 11);
// note no need to worry about remaining chunks - no ambiguity since 11 > 8
int[] arr = new int[len];
int bitOffset = 0, index = 0;
for(int i = 0; i < source.Length; i++)
{
int val = source[i] << bitOffset;
int bitsLeftInVal = 11 - bitOffset;
if(bitsLeftInVal > 8)
{
arr[index] |= val;
bitOffset += 8;
}
else if(bitsLeftInVal == 8)
{
arr[index++] |= val;
bitOffset = 0;
}
else
{
arr[index++] |= (val & 2047);
if(index != arr.Length) arr[index] = val >> 11;
bitOffset = 8 - bitsLeftInVal;
}
}
return arr;
}
}
If you need a different layout you'll need to tweak it.
This encodes 512 MiB in just over a second on my machine.
Overview to the Encode method:
The first thing is does is pre-calculate the amount of space that is going to be required, and allocate the output buffer; since each input contributes 11 bits to the output, this is just some modulo math:
long bits = source.Length * 11;
int len = (int)(bits / 8);
if ((bits % 8) != 0) len++;
byte[] arr = new byte[len];
We know the output position won't match the input, and we know we're going to be starting each 11-bit chunk at different positions in bytes each time, so allocate variables for those, and loop over the input:
int bitOffset = 0, index = 0;
for (int i = 0; i < source.Length; i++)
{
...
}
return arr;
So: taking each input in turn (where the input is the value at position i), take the low 11 bits of the value - and observe that we have 11 bits (of this value) still to write:
int val = source[i] & 2047;
int bitsLeft = 11;
Now, if the current output value is partially written (i.e. bitOffset != 0), we should deal with that first. The amount of space left in the current output is 8 - bitOffset. Since we always have 11 input bits we don't need to worry about having more space than values to fill, so: left-shift our value by bitOffset (pads on the right with bitOffset zeros, as a binary operation), and "or" the lowest 8 bits of this with the output byte. Essentially this says "if bitOffset is 3, write the 5 low bits of val into the 5 high bits of the output buffer"; finally, fixup the values: increment our write position, record that we have fewer bits of the current value still to write, and use right-shift to discard the 8 low bits of val (which is made of bitOffset zeros and 8 - bitOffset "real" bits):
if(bitOffset != 0)
{
val = val << bitOffset;
arr[index++] |= (byte)val;
bitsLeft -= (8 - bitOffset);
val >>= 8;
}
The next question is: do we have (at least) an entire byte of data left? We might not, if bitOffset was 1 for example (so we'll have written 7 bits already, leaving just 4). If we do, we can just stamp that down and increment the write position - then once again track how many are left and throw away the low 8 bits:
if(bitsLeft >= 8)
{
arr[index++] = (byte)val;
bitsLeft -= 8;
val >>= 8;
}
And it is possible that we've still got some left-over; for example, if bitOffset was 7 we'll have written 1 bit in the first chunk, 8 bits in the second, leaving 2 more to write - or if bitOffset was 0 we won't have written anything in the first chunk, 8 in the second, leaving 3 left to write. So, stamp down whatever is left, but do not increment the write position - we've written to the low bits, but the next value might need to write to the high bits. Finally, update bitOffset to be however many low bits we wrote in the last step (which could be zero):
if(bitsLeft != 0)
{
arr[index] = (byte)val;
}
bitOffset = bitsLeft;
The Decode operation is the reverse of this logic - again, calculate the sizes and prepare the state:
int bits = source.Length * 8;
int len = (int)(bits / 11);
int[] arr = new int[len];
int bitOffset = 0, index = 0;
Now loop over the input:
for(int i = 0; i < source.Length; i++)
{
...
}
return arr;
Now, bitOffset is the start position that we want to write to in the current 11-bit value, so if we start at the start, it will be 0 on the first byte, then 8; 3 bits of the second byte join with the first 11-bit integer, so the 5 bits become part of the second - so bitOffset is 5 on the 3rd byte, etc. We can calculate the number of bits left in the current integer by subtracting from 11:
int val = source[i] << bitOffset;
int bitsLeftInVal = 11 - bitOffset;
Now we have 3 possible scenarios:
1) if we have more than 8 bits left in the current value, we can stamp down our input (as a bitwise "or") but do not increment the write position (as we have more to write for this value), and note that we're 8-bits further along:
if(bitsLeftInVal > 8)
{
arr[index] |= val;
bitOffset += 8;
}
2) if we have exactly 8 bits left in the current value, we can stamp down our input (as a bitwise "or") and increment the write position; the next loop can start at zero:
else if(bitsLeftInVal == 8)
{
arr[index++] |= val;
bitOffset = 0;
}
3) otherwise, we have less than 8 bits left in the current value; so we need to write the first bitsLeftInVal bits to the current output position (incrementing the output position), and whatever is left to the next output position. Since we already left-shifted by bitOffset, what this really means is simply: stamp down (as a bitwise "or") the low 11 bits (val & 2047) to the current position, and whatever is left (val >> 11) to the next if that wouldn't exceed our output buffer (padding zeros). Then calculate our new bitOffset:
else
{
arr[index++] |= (val & 2047);
if(index != arr.Length) arr[index] = val >> 11;
bitOffset = 8 - bitsLeftInVal;
}
And that's basically it. Lots of bitwise operations - shifts (<< / >>), masks (&) and combinations (|).

If you wanted to store the least significant 11 bits of an int into two bytes such that the least significant byte has bits 1-8 inclusive and the most significant byte has 9-11:
int toStore = 123456789;
byte msb = (byte) ((toStore >> 8) & 7); //or 0b111
byte lsb = (byte) (toStore & 255); //or 0b11111111
To check this, 123456789 in binary is:
0b111010110111100110100010101
MMMLLLLLLLL
The bits above L are lsb, and have a value of 21, above M are msb and have a value of 5
Doing the work is the shift operator >> where all the binary digits are slid to the right 8 places (8 of them disappear from the right hand side - they're gone, into oblivion):
0b111010110111100110100010101 >> 8 =
0b1110101101111001101
And the mask operator & (the mask operator works by only keeping bits where, in each position, they're 1 in the value and also 1 in the mask) :
0b111010110111100110100010101 &
0b000000000000000000011111111 (255) =
0b000000000000000000000010101
If you're processing an int array, just do this in a loop:
byte[] bs = new byte[ intarray.Length*2 ];
for(int x = 0, b=0; x < intarray.Length; x++){
int toStore = intarray[x];
bs[b++] = (byte) ((toStore >> 8) & 7);
bs[b++] = (byte) (toStore & 255);
}

adjusting image brightness with slider [duplicate]

For windows phone app, when I am adjusting brightness by slider it works fine when I
move it to right. But when I go back to previous position, instead of image darkening, it goes brighter and brighter. Here is my code based on pixel manipulation.
private void slider1_ValueChanged(object sender, RoutedPropertyChangedEventArgs<double> e)
{
wrBmp = new WriteableBitmap(Image1, null);
for (int i = 0; i < wrBmp.Pixels.Count(); i++)
{
int pixel = wrBmp.Pixels[i];
int B = (int)(pixel & 0xFF); pixel >>= 8;
int G = (int)(pixel & 0xFF); pixel >>= 8;
int R = (int)(pixel & 0xFF); pixel >>= 8;
int A = (int)(pixel);
B += (int)slider1.Value; R += (int)slider1.Value; G += (int)slider1.Value;
if (R > 255) R = 255; if (G > 255) G = 255; if (B > 255) B = 255;
if (R < 0) R = 0; if (G < 0) G = 0; if (B < 0) B = 0;
wrBmp.Pixels[i] = B | (G << 8) | (R << 16) | (A << 24);
}
wrBmp.Invalidate();
Image1.Source = wrBmp;
}
What am I missing and is there any problem with slider value. I am working with small images as usual in mobiles. I have already tried copying original image to duplicate one. I think code is perfect, after a lot of research I found that the problem is due to slider value.Possible solution is assigning initial value to slider. I want some code help.
private double lastSlider3Vlaue;
private void slider3_ValueChanged(object sender,`RoutedPropertyChangedEventArgs e)
{
if (slider3 == null) return;
double[] contrastArray = { 1, 1.2, 1.3, 1.6, 1.7, 1.9, 2.1, 2.4, 2.6, 2.9 };
double CFactor = 0;
int nIndex = 0;
nIndex = (int)slider3.Value - (int)lastSlider3Vlaue;
if (nIndex < 0)
{
nIndex = (int)lastSlider3Vlaue - (int)slider3.Value;
this.lastSlider3Vlaue = slider3.Value;
CFactor = contrastArray[nIndex];
}
else
{
nIndex = (int)slider3.Value - (int)lastSlider3Vlaue;
this.lastSlider3Vlaue = slider3.Value;
CFactor = contrastArray[nIndex];
}
WriteableBitmap wbOriginal;
wbOriginal = new WriteableBitmap(Image1, null);
wrBmp = new WriteableBitmap(wbOriginal.PixelWidth, wbOriginal.PixelHeight);
wbOriginal.Pixels.CopyTo(wrBmp.Pixels, 0);
int h = wrBmp.PixelHeight;
int w = wrBmp.PixelWidth;
for (int i = 0; i < wrBmp.Pixels.Count(); i++)
{
int pixel = wrBmp.Pixels[i];
int B = (int)(pixel & 0xFF); pixel >>= 8;
int G = (int)(pixel & 0xFF); pixel >>= 8;
int R = (int)(pixel & 0xFF); pixel >>= 8;
int A = (int)(pixel);
R = (int)(((R - 128) * CFactor) + 128);
G = (int)(((G - 128) * CFactor) + 128);
B = (int)(((B - 128) * CFactor) + 128);
if (R > 255) R = 255; if (G > 255) G = 255; if (B > 255) B = 255;
if (R < 0) R = 0; if (G < 0) G = 0; if (B < 0) B = 0;
wrBmp.Pixels[i] = B | (G << 8) | (R << 16) | (A << 24);
}
wrBmp.Invalidate();
Image1.Source = wrBmp;
}
After debugging I found that the r g b values are decreasing continuosly when sliding forward,but when sliding backwards it is also decreasing where as it shoul increase.
please help iam working on this since past last three months.Besides this u also give me advice about how i can complete this whole image processing

Your algorithm is wrong. Each time the slider's value changes, you're adding that value to the picture's brightness. What makes your logic flawed is that the value returned by the slider will always be positive, and you're always adding the brightness to the same picture.
So, if the slider starts with a value of 10, I'll add 10 to the picture's brightness.
Then, I slide to 5. I'll add 5 to the previous picture's brightness (the one you already added 10 of brightness to).
Two ways to solve the issue:
Keep a copy of the original picture, and duplicate it every time your method is called. Then add the brightness to the copy (and not the original). That's the safest way.
Instead of adding the new absolute value of the slider, calculate the relative value (how much it changed since the last time the method was called:
private double lastSliderValue;
private void slider1_ValueChanged(object sender, RoutedPropertyChangedEventArgs<double> e)
{
var offset = slider1.Value - this.lastSliderValue;
this.lastSliderValue = slider1.Value;
// Insert your old algorithm here, but replace occurences of "slider1.Value" by "offset"
}
This second way can cause a few headaches though. Your algorithm is capping the RGB values to 255. In those cases, you are losing information and cannot revert back to the old state. For instance, take the extreme example of a slider value of 255. The algorithm sets all the pixels to 255, thus generating a white picture. Then you reduce the slider to 0, which should in theory restore the original picture. In this case, you'll subtract 255 to each pixel, but since every pixel's value is 255 you'll end up with a black picture.
Therefore, except if you find a clever way to solve the issue mentionned in the second solution, I'd recommand going with the first one.

How to reverse that function

I've asked before about the opposite of Bitwise AND(&) and you told me its impossible to reverse.
Well,this is the situation: The server sends an image,which is encoded with the function I want to reverse,then it is encoded with zlib.
This is how I get the image from the server:
UInt32[] image = new UInt32[200 * 64];
int imgIndex = 0;
byte[] imgdata = new byte[compressed];
byte[] imgdataout = new byte[uncompressed];
Array.Copy(data, 17, imgdata, 0, compressed);
imgdataout = zlib.Decompress(imgdata);
for (int h = 0; h < height; h++)
{
for (int w = 0; w < width; w++)
{
imgIndex = (int)((height - 1 - h) * width + w);
image[imgIndex] = 0xFF000000;
if (((1 << (Int32)(0xFF & (w & 0x80000007))) & imgdataout[((h * width + w) >> 3)]) > 0)
{
image[imgIndex] = 0xFFFFFFFF;
}
}
}
Width,Height,Image decompressed and Image compressed length are always the same.
When this function is done I put image(UInt32[] array) in a Bitmap and I've got it.
Now I want to be the server and send that image.I have to do two things:
Reverse that function and then compress it with zlib.
How do I reverse that function so I can encode the picture?
for (int h = 0; h < height; h++)
{
for (int w = 0; w < width; w++)
{
imgIndex = (int)((height - 1 - h) * width + w);
image[imgIndex] = 0xFF000000;
if (((1 << (Int32)(0xFF & (w & 0x80000007))) & imgdataout[((h * width + w) >> 3)]) > 0)
{
image[imgIndex] = 0xFFFFFFFF;
}
}
}
EDIT:The format is 32bppRGB

The assumption that the & operator is always irreversible is incorrect.
Yes, in general if you have
c = a & b
and all you know is the value of c, then you cannot know what values a or b had before hand.
However it's very common for & to be used to extract certain bits from a longer value, where those bits were previously combined together with the | operator and where each 'bit field' is independent of every other. The fundamental difference with the generic & or | operators that makes this reversible is that the original bits were all zero beforehand, and the other bits in the word are left unchanged. i.e:
0xc0 | 0x03 = 0xc3 // combine two nybbles
0xc3 & 0xf0 = 0xc0 // extract the top nybble
0xc3 & 0x0f = 0x03 // extract the bottom nybble
In this case your current function appears to be extracting a 1 bit-per-pixel (monochrome image) and converting it to 32-bit RGBA.
You'll need something like:
int source_image[];
byte dest_image[];
for (int h = 0; h < height; ++h) {
for (int w = 0; w < width; ++w) {
int offset = (h * width) + w;
if (source_image[offset] == 0xffffffff) {
int mask = w % 8; // these two lines convert from one int-per-pixel
offset /= 8; // offset to one-bit-per-pixel
dest_image[offset] |= (1 << mask); // only changes _one_ bit
}
}
}
NB: assumes the image is a multiple of 8 pixels wide, that the dest_image array was previously all zeroes. I've used % and / in that inner test because it's easier to understand and the compiler should convert to mask / shift itself. Normally I'd do the masking and shifting myself.