I got a 16 bpp bitmap that I converted to 32 bpp via code below:
void Rgb555ToRgb8(const UChar* bitmapData, UInt32 width, UInt32 height, UChar* buf)
{
UInt32 dst_bytes_per_row = width * 4;
UInt32 src_bytes_per_row = ((width * 16 + 31) / 32) * 4;
UInt16 red_mask = 0x7C00;
UInt16 green_mask = 0x3E0;
UInt16 blue_mask = 0x1F;
for (UInt32 row = 0; row < height; ++row)
{
UInt32 dstCol = 0, srcCol = 0;
do
{
UInt16 rgb = *(UInt16*)(bitmapData + row * src_bytes_per_row + srcCol);
UChar red_value = (rgb & red_mask) >> 10;
UChar green_value = (rgb & green_mask) >> 5;
UChar blue_value = (rgb & blue_mask);
buf[row*dst_bytes_per_row + dstCol] = blue_value << 3;
buf[row*dst_bytes_per_row + dstCol + 1] = green_value << 3;
buf[row*dst_bytes_per_row + dstCol + 2] = red_value << 3;
buf[row*dst_bytes_per_row + dstCol + 3] = rgb >> 15;
srcCol += 2;
dstCol += 4;
} while (srcCol < src_bytes_per_row);
}
}
Here is conversion result: [2]: https://i.stack.imgur.com/1ajO7.png
I also tried to convert this image via GdiPlus:
Gdiplus::Bitmap* bmp = new Gdiplus::Bitmap(w,h,PixelFormat32bppRGB);
Resultant image is .
Notice that the 2 results don't look exactly the same (e.g., the background in GdiPlus result is white). How can I modify my code to match GdiPlus result?
There are two issues that need to be addressed:
Unused bits
When moving from 5 bits of information to 8 bits of information you gain an additional 3 bits. As implemented, the code doesn't make use of that additional range, and is biased towards darker color components. This is an illustration of what blue_value << 3 actually does:
5 bits per channel 8 bits per channel
bbbbb -> bbbbb000
To address this, the least significant 3 bits need to grow as the channel value gets higher. A simple (yet somewhat inaccurate) would be to just copy the most significant 3 bits down to the least significant 3 bits, i.e.
buf[row*dst_bytes_per_row + dstCol] = (blue_value << 3) | (blue_value >> 2);
buf[row*dst_bytes_per_row + dstCol + 1] = (green_value << 3) | (green_value >> 2);
buf[row*dst_bytes_per_row + dstCol + 2] = (red_value << 3) | (red_value >> 2);
The exact mapping would be a bit more involved, something like
blue_value = static_cast<UChar>((blue_value * 255.0) / 31.0 + 0.5);
That converts from 5 bits to the respective 8 bit value that's nearest to the ideal value, including the 4 values that were 1/255th off in the bit-shifting solution above.
If you opt for the latter, you can build a lookup table that stores the mapped values. This table is only 32 entries of one byte each, so it fits into a single cache-line.
Alpha channel
Assuming that the MSB of your source image is indeed interpreted as an alpha value, you're going to have move that into the destination as well. Since the source is only 1 bit of information, the raw transformation is trivial:
buf[row*dst_bytes_per_row + dstCol + 3] = rgb & (1 << 15) ? 255 : 0;
That may or may not be all that's needed. Windows assumes premultiplied alpha, i.e. the stored values of the color channels must be premultiplied by the alpha value (see BLENDFUNCTION for reference).
If the alpha value is 255, the color channel values are already correct. If the alpha value is 0, all color channels need to be multiplied by zero (or simply set to 0). The translation doesn't produce any other alpha values.
Related
I am trying to covert 16bit (RBG565) to grayscale image. I tried various combination of formula suggested in the internet all works good for 24 bit RGB888 format. when i try with 16bit (RBG565) the image has blue, red pixels, unable to create the exact grayscale image. please help.
Formula 1 works better than Formula 2:
Formula 1:
unsigned char gray = (red * 77+( (green )* 150)/2 + blue * 29+128) / 256;
Formula 2:
unsigned char gray = red * 0.212 + green * 0.715 + blue * 0.072;
Since the RGB components tend to be a scale from no-colour to full-colour for each primary colour (primary in the general sense, meaning colours combined to form others), I would probably convert to RGB up front, then use the normal grey-scaling algorithm (any of the ones you state that "work good for 24 bit RGB888").
That first bit could probably be done with (upper case is the eight-bit value, lower case the 5/6-bit value):
R = r * 8 ; 256/32, 8-bit from 5-bit
B = b * 4 ; 256/64, 8-bit from 6-bit
G = g * 8 ; 256/32, 8-bit from 5-bit
Just ensure you get the order right when extracting the bits, because you're converting RBG to RGB.
For example here's a complete C program (since you didn't provide a language tag) containing a function which will do that, except that it uses slightly different scaling which relies on percentages rather than fixed values that need 0..255 inputs:
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
uint16_t greyScale(uint16_t rbg565) {
// Get the values (32-bit to hold larger intermediate values).
uint32_t r = (rbg565 >> 11) & 31U; // XXXXX___________
uint32_t g = rbg565 & 31U; // ___________XXXXX
uint32_t b = (rbg565 >> 5) & 63U; // _____XXXXXX_____
// Scale each of them up to the range 0..500.
g = g * 500 / 31;
b = b * 500 / 63;
r = r * 500 / 31;
// Use example RGB weights of 299/587/114 (summing to 1,000).
// Range then 0..500,000 so divide by 500 to get grey 0..1,000.
uint32_t grey = (r * 299 + g * 587 + b * 114);
grey /= 500;
return (uint16_t)grey;
}
int main(int argc, char *argv[])
{
for (int i = 1; i < argc; i++) {
uint16_t rbg = atoi(argv[i]);
printf("%5d -> %d\n", rbg, greyScale(rbg));
}
return 0;
}
If you run it with some sample values, you'll see the greyscale values come out (comments are added by me):
pax:~> ./prog 0 65535 31 $((63*32)) $((31*2048)) $((16*2048+32*32+16))
0 -> 0 # No colour components.
65535 -> 1000 # Max RGB (white).
31 -> 587 # Max green, nothing else.
2016 -> 114 # Max blue, nothing else.
63488 -> 299 # Max red, nothing else.
33808 -> 514 # A little more than half of everything.
If you want to use weightings different from the ones I provided (29.9%, 58.7%, and 11.4%), just change this line to match, making sure they still sum to a thousand:
uint32_t grey = (r * 299 + g * 587 + b * 114);
I need to calculate the mean of a 3D matrices (last step in the code). However, there are many NaNs in the (diff_dataframe./dataframe_vor) calculation. So when I use this code, some results will be NaN. How could I calculate the mean of this matrix by ignoring the NaNs? I attached the code as below.
S.amplitude = 1:20;%:20;
S.blocksize = [1 2 3 4 5 6 8 10 12 15 20];
S.frameWidth = 1920;
S.frameHeight = 1080;
S.quality=0:10:100;
image = 127*ones(S.frameHeight,S.frameWidth,3);
S.yuv2rgb = [1 0 1.28033; 1 -0.21482 -0.38059; 1 2.12798 0];
i_bs = 0;
for BS = S.blocksize
i_bs = i_bs + 1;
hblocks = S.frameWidth / BS;
vblocks = S.frameHeight / BS;
i_a = 0;
dataU = randi([0 1],vblocks,hblocks);
dataV = randi([0 1],vblocks,hblocks);
dataframe_yuv = zeros(S.frameHeight, S.frameWidth, 3);
for x = 1 : hblocks
for y = 1 : vblocks
dataframe_yuv((y-1)*BS+1:y*BS, ...
(x-1)*BS+1:x*BS, 2) = dataU(y,x) * 2 - 1;
dataframe_yuv((y-1)*BS+1:y*BS, ...
(x-1)*BS+1:x*BS, 3) = dataV(y,x) * 2 - 1;
end
end
dataframe_rgb(:,:,1) = S.yuv2rgb(1,1) * dataframe_yuv(:,:,1) + ...
S.yuv2rgb(1,2) * dataframe_yuv(:,:,2) + ...
S.yuv2rgb(1,3) * dataframe_yuv(:,:,3);
dataframe_rgb(:,:,2) = S.yuv2rgb(2,1) * dataframe_yuv(:,:,1) + ...
S.yuv2rgb(2,2) * dataframe_yuv(:,:,2) + ...
S.yuv2rgb(2,3) * dataframe_yuv(:,:,3);
dataframe_rgb(:,:,3) = S.yuv2rgb(3,1) * dataframe_yuv(:,:,1) + ...
S.yuv2rgb(3,2) * dataframe_yuv(:,:,2) + ...
S.yuv2rgb(3,3) * dataframe_yuv(:,:,3);
for A = S.amplitude
i_a = i_a + 1;
i_q = 0;
image1p = round(image + dataframe_rgb * A);
image1n = round(image - dataframe_rgb * A);
dataframe_vor = ((image1p-image1n)/2)/255;
for Q = S.quality
i_q = i_q + 1;
namestrp = ['greyjpegs/Img_BS' num2str(BS) '_A' num2str(A) '_Q' num2str(Q) '_1p.jpg'];
namestrn = ['greyjpegs/Img_BS' num2str(BS) '_A' num2str(A) '_Q' num2str(Q) '_1n.jpg'];
imwrite(image1p/255,namestrp,'jpg', 'Quality', Q);
imwrite(image1n/255,namestrn,'jpg', 'Quality', Q);
error_mean(i_bs, i_a, i_q) = mean2((abs(diff_dataframe./dataframe_vor)));
end
end
end
mean2 is a shortcut function that's part of the image processing toolbox that finds the entire average of a 2D region which doesn't include handling NaN. In that case, simply remove all values that are NaN and find the resulting average. Note that the removal of NaN unrolls the 2D region into a 1D vector, so we can simply use mean in this case. As an additional check, let's make sure there are no divide by 0 errors, so also check for Inf as well.
Therefore, replace this line:
error_mean(i_bs, i_a, i_q) = mean2((abs(diff_dataframe./dataframe_vor)));
... with:
tmp = abs(diff_dataframe ./ dataframe_vor);
mask = ~isnan(tmp) | ~isinf(tmp);
tmp = tmp(mask);
if isempty(tmp)
error_mean(i_bs, i_a, i_q) = 0;
else
error_mean(i_bs, i_a, i_q) = mean(tmp);
We first assign the desired operation to a temporary variable, use isnan and isinf to remove out the offending values, then find the average of the rest. One intricacy is that if your entire region is NaN or Inf, then the removal of all these entries in the region results in the empty vector, and finding the mean of this undefined. A separate check is there to be sure that if it's empty, simply assign the value of 0 instead.
I'm sure this is fairly simple, however I have a major mental block on it, so I need a little help here!
I have an array of 5 integers, the array is already filled with some data. I want to set the last N bits of the array to be random noise.
[int][int][int][int][int]
eg. set last 40 bits
[unchanged][unchanged][unchanged][24 bits of old data followed 8 bits of randomness][all random]
This is largely language agnostic, but I'm working in C# so bonus points for answers in C#
Without any bit-fu in C#:
BitArray ba = new BitArray (originalIntArray);
for (int i = startToReplaceFrom; i < endToReplaceTo; i++)
ba.Set (i, randomValue);
When you XOR any data with random data, the result is random, so you can do this:
Random random = new Random();
x[x.Length - 2] ^= random.Next(1 << 8);
x[x.Length - 1] = random.Next(1 << 16);
x[x.Length - 1] ^= random.Next(1 << 16) << 16;
For a general solution for any N you can use a loop:
for (int i = 0; i < 40; ++i)
{
x[x.Length - i / 32 - 1] ^= random.Next(2) << (i % 32);
}
Note that this calls random more times than necessary, but is simple.
In pseudo-Python:
N = 5 # array size
bits = 40 # for instance
int_bits = 32 # bits in one integer
i = N
while bits > 0:
value_bits = min (bits, int_bits)
bits -= value_bits
mask = (1 << value_bits) - 1
i -= 1
array[i] ^= random () & mask
Int32 is 4 bytes or 32 bits.
So you need the last int, and 8 bits extra.
int lastEightBitsMask = 0x000F + 1;
Random rand = new Random();
arr[arr.Length - 1] = rand.Next();
arr[arr.Length - 2] ^= rand.Next(lastEightBitsMask);
Explanation:
The last element's modification should be pretty clear - if you need the last 40 bits, the last 32 bits are included in that.
The remaining eight bits's modification is bounded above by 0x000F + 1, since rand.Next's argument is an exclusive upper bound, the randoms generated will be no more than that. The remaining bits of the number will stay the same, since 1^0 == 1 and 0^0 == 0.
I’ve got my hands on a 16-bit rgb565 image (specifically, an Android framebuffer dump), and I would like to convert it to 24-bit rgb888 for viewing on a normal monitor.
The question is, how does one convert a 5- or 6-bit channel to 8 bits? The obvious answer is to shift it. I started out by writing this:
puts("P6 320 480 255");
uint16_t buf;
while (read(0, &buf, sizeof buf)) {
unsigned char red = (buf & 0xf800) >> 11;
unsigned char green = (buf & 0x07e0) >> 5;
unsigned char blue = buf & 0x001f;
putchar(red << 3);
putchar(green << 2);
putchar(blue << 3);
}
However, this doesn’t have one property I would like, which is for 0xffff to map to 0xffffff, instead of 0xf8fcf8. I need to expand the value in some way, but I’m not sure how that should work.
The Android SDK comes with a tool called ddms (Dalvik Debug Monitor) that takes screen captures. As far as I can tell from reading the code, it implements the same logic; yet its screenshots are coming out different, and white is mapping to white.
Here’s the raw framebuffer, the smart conversion by ddms, and the dumb conversion by the above algorithm. Note that the latter is slightly darker and greener.
(By the way, this conversion is implemented in ffmpeg, but it’s just performing the dumb conversion listed above, leaving the LSBs at all zero.)
I guess I have two questions:
What’s the most sensible way to convert rgb565 to rgb888?
How is DDMS converting its screenshots?
You want to map each of these from a 5/6 bit space to an 8 bit space.
5 bits = 32 values
6 bits = 64 values
8 bits = 256 values
The code you're using is taking the naive approach that x5 * 256/32 = x8 where 256/32 = 8 and multiplying by 8 is left shift 3 but, as you say, this doesn't necessarily fill the new number space "correctly". 5 to 8 for max value is 31 to 255 and therein lies your clue to the solution.
x8 = 255/31 * x5
x8 = 255/63 * x6
where x5, x6 and x8 are 5, 6 and 8 bit values respectively.
Now there is a question about the best way to implement this. It does involve division and with integer division you will lose any remainder result (round down basically) so the best solution is probably to do floating point arithmetic and then round half up back to an integer.
This can be sped up considerably by simply using this formula to generate a lookup table for each of the 5 and 6 bit conversions.
My few cents:
If you care about precise mapping, yet fast algorithm you can consider this:
R8 = ( R5 * 527 + 23 ) >> 6;
G8 = ( G6 * 259 + 33 ) >> 6;
B8 = ( B5 * 527 + 23 ) >> 6;
It uses only: MUL, ADD and SHR -> so it is pretty fast!
From the other side it is compatible in 100% to floating point mapping with proper rounding:
// R8 = (int) floor( R5 * 255.0 / 31.0 + 0.5);
// G8 = (int) floor( G6 * 255.0 / 63.0 + 0.5);
// B8 = (int) floor( R5 * 255.0 / 31.0 + 0.5);
Some extra cents:
If you are interested in 888 to 565 conversion, this works very well too:
R5 = ( R8 * 249 + 1014 ) >> 11;
G6 = ( G8 * 253 + 505 ) >> 10;
B5 = ( B8 * 249 + 1014 ) >> 11;
Constants were found using brute force search with somę early rejections to speed thing up a bit.
You could shift and then or with the most significant bits; i.e.
Red 10101 becomes 10101000 | 101 => 10101101
12345 12345--- 123 12345123
This has the property you seek, but it's not the most linear mapping of values from one space to the other. It's fast, though. :)
Cletus' answer is more complete and probably better. :)
iOS vImage Conversion
The iOS Accelerate Framework documents the following algorithm for the vImageConvert_RGB565toARGB8888 function:
Pixel8 alpha = alpha
Pixel8 red = (5bitRedChannel * 255 + 15) / 31
Pixel8 green = (6bitGreenChannel * 255 + 31) / 63
Pixel8 blue = (5bitBlueChannel * 255 + 15) / 31
For a one-off conversion this will be fast enough, but if you want to process many frames you want to use something like the iOS vImage conversion or implement this yourself using NEON intrinsics.
From ARMs Community Forum Tutorial
First, we will look at converting RGB565 to RGB888. We assume there are eight 16-bit pixels in register q0, and we would like to separate reds, greens and blues into 8-bit elements across three registers d2 to d4.
vshr.u8 q1, q0, #3 # shift red elements right by three bits,
# discarding the green bits at the bottom of
# the red 8-bit elements.
vshrn.i16 d2, q1, #5 # shift red elements right and narrow,
# discarding the blue and green bits.
vshrn.i16 d3, q0, #5 # shift green elements right and narrow,
# discarding the blue bits and some red bits
# due to narrowing.
vshl.i8 d3, d3, #2 # shift green elements left, discarding the
# remaining red bits, and placing green bits
# in the correct place.
vshl.i16 q0, q0, #3 # shift blue elements left to most-significant
# bits of 8-bit color channel.
vmovn.i16 d4, q0 # remove remaining red and green bits by
# narrowing to 8 bits.
The effects of each instruction are described in the comments above, but in summary, the operation performed on each channel is:
Remove color data for adjacent channels using shifts to push the bits off either end of the element.
Use a second shift to position the color data in the most-significant bits of each element, and narrow to reduce element size from 16 to eight bits.
Note the use of element sizes in this sequence to address 8 and 16 bit elements, in order to achieve some of the masking operations.
A small problem
You may notice that, if you use the code above to convert to RGB888 format, your whites aren't quite white. This is because, for each channel, the lowest two or three bits are zero, rather than one; a white represented in RGB565 as (0x1F, 0x3F, 0x1F) becomes (0xF8, 0xFC, 0xF8) in RGB888. This can be fixed using shift with insert to place some of the most-significant bits into the lower bits.
For an Android specific example I found a YUV-to-RGB conversion written in intrinsics.
Try this:
red5 = (buf & 0xF800) >> 11;
red8 = (red5 << 3) | (red5 >> 2);
This will map all zeros into all zeros, all 1's into all 1's, and everything in between into everything in between. You can make it more efficient by shifting the bits into place in one step:
redmask = (buf & 0xF800);
rgb888 = (redmask << 8) | ((redmask<<3)&0x070000) | /* green, blue */
Do likewise for green and blue (for 6 bits, shift left 2 and right 4 respectively in the top method).
The general solution is to treat the numbers as binary fractions - thus, the 6 bit number 63/63 is the same as the 8 bit number 255/255. You can calculate this using floating point math initially, then compute a lookup table, as other posters suggest. This also has the advantage of being more intuitive than bit-bashing solutions. :)
There is an error jleedev !!!
unsigned char green = (buf & 0x07c0) >> 5;
unsigned char blue = buf & 0x003f;
the good code
unsigned char green = (buf & 0x07e0) >> 5;
unsigned char blue = buf & 0x001f;
Cheers,
Andy
I used the following and got good results. Turned out my Logitek cam was 16bit RGB555 and using the following to convert to 24bit RGB888 allowed me to save as a jpeg using the smaller animals ijg: Thanks for the hint found here on stackoverflow.
// Convert a 16 bit inbuf array to a 24 bit outbuf array
BOOL JpegFile::ByteConvert(BYTE* inbuf, BYTE* outbuf, UINT width, UINT height)
{ UINT row_cnt, pix_cnt;
ULONG off1 = 0, off2 = 0;
BYTE tbi1, tbi2, R5, G5, B5, R8, G8, B8;
if (inbuf==NULL)
return FALSE;
for (row_cnt = 0; row_cnt <= height; row_cnt++)
{ off1 = row_cnt * width * 2;
off2 = row_cnt * width * 3;
for(pix_cnt=0; pix_cnt < width; pix_cnt++)
{ tbi1 = inbuf[off1 + (pix_cnt * 2)];
tbi2 = inbuf[off1 + (pix_cnt * 2) + 1];
B5 = tbi1 & 0x1F;
G5 = (((tbi1 & 0xE0) >> 5) | ((tbi2 & 0x03) << 3)) & 0x1F;
R5 = (tbi2 >> 2) & 0x1F;
R8 = ( R5 * 527 + 23 ) >> 6;
G8 = ( G5 * 527 + 23 ) >> 6;
B8 = ( B5 * 527 + 23 ) >> 6;
outbuf[off2 + (pix_cnt * 3)] = R8;
outbuf[off2 + (pix_cnt * 3) + 1] = G8;
outbuf[off2 + (pix_cnt * 3) + 2] = B8;
}
}
return TRUE;
}
Here's the code:
namespace convert565888
{
inline uvec4_t const _c0{ { { 527u, 259u, 527u, 1u } } };
inline uvec4_t const _c1{ { { 23u, 33u, 23u, 0u } } };
} // end ns
uvec4_v const __vectorcall rgb565_to_888(uvec4_v const rgba) {
return(uvec4_v(_mm_srli_epi32(_mm_add_epi32(_mm_mullo_epi32(rgba.v,
uvec4_v(convert565888::_c0).v), uvec4_v(convert565888::_c1).v), 6)));
}
and for rgb 888 to 565 conversion:
namespace convert888565
{
inline uvec4_t const _c0{ { { 249u, 509u, 249u, 1u } } };
inline uvec4_t const _c1{ { { 1014u, 253u, 1014u, 0u } } };
} // end ns
uvec4_v const __vectorcall rgb888_to_565(uvec4_v const rgba) {
return(uvec4_v(_mm_srli_epi32(_mm_add_epi32(_mm_mullo_epi32(rgba.v,
uvec4_v(convert888565::_c0).v), uvec4_v(convert888565::_c1).v), 11)));
}
for the explanation of where all these numbers come from, specifically how I calculated the optimal multiplier and bias for green:
Desmos graph -
https://www.desmos.com/calculator/3grykboay1
The graph isn't the greatest but it shows the actual value vs. error -- play around with the interactive sliders to see how different values affect the output. This graph also applies to calculating the red and blue values aswell. Typically green is shifted by 10bits, red and blue 11bits.
In order for this to work with intrinsic _mm_srli_epi32 / _mm_srl_epi32 requires all components to be shifted by the same amount. So everything is shifted by 11 bits (rgb888_to_565) in this version, however, the green component is scaled to compensate for this change. Fortunately, it scales perfectly!
I had this difficulty too, and the most faithful way I found was to replace the 16-bit value with the original 24-bit value. Now the ILI9341 screen color is visually compatible with Notebook screen. I thought of just using the 24-bit color table, but then the display routines would have to be converted to 565, and that would make the program even slower.
If the color palette is fixed as in my case, it might be the most viable option. I tried to make use of the 3 MSB adding with the 3 LSB, but it wasn't very good.
The colors I used on the ILI9341 display I got from this website (Note: I choose the 24-bit color 888 and get the 16-bit color 565, on this website there's no way to do otherwise):
http://www.barth-dev.de/online/rgb565-color-picker/
For example, I read the pixel color of the ILI9341 display and save it to a USB Disk, in a file, in BMP format. As the display operates with 16-bit or 18-bit, I have no way to retrieve 24-bit information directly from the GRAM memory.
#define BLACK_565 0x0000
#define BLUE_565 0x001F
#define RED_565 0xF800
#define GREEN_565 0x07E0
#define CYAN_565 0x07FF
#define MAGENTA_565 0xF81F
#define YELLOW_565 0xFFE0
#define WHITE_565 0xFFFF
#define LIGHTGREY_565 0xC618
#define ORANGE_565 0xFD20
#define GREY_565 0x8410
#define DARKGREY_565 0x2104
#define DARKBLUE_565 0x0010
#define DARKGREEN_565 0x03E0
#define DARKCYAN_565 0x03EF
#define DARKYELLOW_565 0x8C40
#define BLUESKY_565 0x047F
#define BROWN_565 0xC408
#define BLACK_888 0x000000
#define BLUE_888 0x0000FF
#define RED_888 0xFF0000
#define GREEN_888 0x04FF00
#define CYAN_888 0x00FFFB
#define MAGENTA_888 0xFF00FA
#define YELLOW_888 0xFBFF00
#define WHITE_888 0xFFFFFF
#define LIGHTGREY_888 0xC6C3C6
#define ORANGE_888 0xFFA500
#define GREY_888 0x808080
#define DARKGREY_888 0x202020
#define DARKBLUE_888 0x000080
#define DARKGREEN_888 0x007D00
#define DARKCYAN_888 0x007D7B
#define DARKYELLOW_888 0x898A00
#define BLUESKY_888 0x008CFF
#define BROWN_888 0xC08240
I did the test (using an STM32F407 uC) with an IF statement, but it can also be done with Select Case, or another form of comparison.
uint16_t buff1; // pixel color value read from GRAM
uint8_t buff2[3];
uint32_t color_buff; // to save to USB disk
if (buff1 == BLUE_565) color_buff = BLUE_888;
else if (buff1 == RED_565) color_buff = RED_888;
else if (buff1 == GREEN_565) color_buff = GREEN_888;
else if (buff1 == CYAN_565) color_buff = CYAN_888;
else if (buff1 == MAGENTA_565) color_buff = MAGENTA_888;
else if (buff1 == YELLOW_565) color_buff = YELLOW_888;
else if (buff1 == WHITE_565) color_buff = WHITE_888;
else if (buff1 == LIGHTGREY_565) color_buff = LIGHTGREY_888;
else if (buff1 == ORANGE_565) color_buff = ORANGE_888;
else if (buff1 == GREY_565) color_buff = GREY_888;
else if (buff1 == DARKGREY_565) color_buff = DARKGREY_888;
else if (buff1 == DARKBLUE_565) color_buff = DARKBLUE_888;
else if (buff1 == DARKCYAN_565) color_buff = DARKCYAN_888;
else if (buff1 == DARKYELLOW_565) color_buff = DARKYELLOW_888;
else if (buff1 == BLUESKY_565) color_buff = BLUESKY_888;
else if (buff1 == BROWN_565) color_buff = BROWN_888;
else color_buff = BLACK;
RGB separation for saving to 8-bit variables:
buff2[0] = color_buff; // Blue
buff2[1] = color_buff >> 8; // Green
buff2[2] = color_buff >> 16; // Red
I have some RGB(image) data which is 12 bit. Each R,G,B has 12 bits, total 36 bits.
Now I need to club this 12 bit RGB data into a packed data format. I have tried to mention the packing as below:-
At present I have input data as -
B0 - 12 bits G0 - 12 bits R0 - 12 bits B1 - 12 bits G1 - 12 bits R1 - 12 bits .. so on.
I need to convert it to packed format as:-
Byte1 - B8 (8 bits of B0 data)
Byte2 - G4B4 (remaining 4 bits of B0 data+ first 4 bits of G0)
Byte3 - G8 (remaining 8 bits of G0)
Byte4 - R8 (first 8 bits of R0)
Byte5 - B4R4 (first 4 bits of B1 + last 4 bits of R0)
I have to write these individual bytes to a file in text format. one byte below another.
Similar thing i have to do for a 10 bit RGB input data.
Is there any tool/software to get the conversion of data i am looking to get done.
I am trying to do it in a C program - I am forming a 64 bit from the individual 12 bits of R,G,B (total 36 bits). But after that I am not able to come up with a logic to pick
the necessary bits from a R,G,B data to form a byte stream, and to dump them to a text file.
Any pointers will be helpful.
This is pretty much untested, super messy code I whipped together to give you a start. It's probably not packing the bytes exactly as you want, but you should get the general idea.
Apologies for the quick and nasty code, only had a couple of minutes, hope it's of some help anyway.
#include <stdio.h>
typedef struct
{
unsigned short B;
unsigned short G;
unsigned short R;
} UnpackedRGB;
UnpackedRGB test[] =
{
{0x0FFF, 0x000, 0x0EEE},
{0x000, 0x0FEF, 0xDEF},
{0xFED, 0xDED, 0xFED},
{0x111, 0x222, 0x333},
{0xA10, 0xB10, 0xC10}
};
UnpackedRGB buffer = {0, 0, 0};
int main(int argc, char** argv)
{
int numSourcePixels = sizeof(test)/sizeof(UnpackedRGB);
/* round up to the last byte */
int destbytes = ((numSourcePixels * 45)+5)/10;
unsigned char* dest = (unsigned char*)malloc(destbytes);
unsigned char* currentDestByte = dest;
UnpackedRGB *pixel1;
UnpackedRGB *pixel2;
int ixSource;
for (ixSource = 0; ixSource < numSourcePixels; ixSource += 2)
{
pixel1 = &test[ixSource];
pixel2 = ((ixSource + 1) < numSourcePixels ? &test[ixSource] : &buffer);
*currentDestByte++ = (0x0FF) & pixel1->B;
*currentDestByte++ = ((0xF00 & pixel1->B) >> 8) | (0x0F & pixel1->G);
*currentDestByte++ = ((0xFF0 & pixel1->G) >> 4);
*currentDestByte++ = (0x0FF & pixel1->R);
*currentDestByte++ = ((0xF00 & pixel1->R) >> 8) | (0x0F & pixel2->B);
if ((ixSource + 1) >= numSourcePixels)
{
break;
}
*currentDestByte++ = ((0xFF0 & pixel2->B) >> 4);
*currentDestByte++ = (0x0FF & pixel2->G);
*currentDestByte++ = ((0xF00 & pixel2->G) >> 8) | (0x0F & pixel2->R);
*currentDestByte++ = (0xFF0 & pixel2->R);
}
FILE* outfile = fopen("output.bin", "w");
fwrite(dest, 1, destbytes,outfile);
fclose(outfile);
}
Use bitwise & (and), | (or), and shift <<, >> operators.