Using BufferedImage to create an image in Java

Question

I am trying re-create an image using given 2D pixel arrays (rows, and columns) using the setRGB() method in BufferedImage.

Below is the follwing code:

BufferedImage bufferedImage = new BufferedImage(reconstructedJPEG[0].length, reconstructedJPEG.length, BufferedImage.TYPE_INT_RGB);
//loop through redPixels[][] array
for(int row=0; row<redPixels.length; row++){
    for(int col=0; col<redPixels[0].length; col++){
        //call setRGB() on redPixels
        bufferedImage.setRGB(col, row, (redPixels[row][col]));
    }
}

The code above works, but I am not sure how I can also set the green and blue pixel arrays? Right now, its a very dull, dark red/purple image, that does no look like the original image.

Also, is there a another away I can form these arrays into a 1D image (which would be its raw pixels, red+green+blue components into one integer?

Thanks any help would be great.

dinox0r · Accepted Answer · 2014-03-04 16:25:24Z

2

Combine the individual color values for the 3 channels (red, green and blue) in one pixel using bitwise operators:

int rgb = (redValue & 0xff) << 16 | (greenValue & 0xff) << 8 | (blueValue & 0xff);

Then call setRGB with the composed value as parameter:

bufferedImage.setRGB(col, row, rgb);

The bitwise operation sentence can be cumbersome at first sight but it does the following:

Take every channel value and make it 8-bit range based value (0, 255) using the & 0xff mask (the format BufferedImage.TYPE_INT_RGB expects channels to be 8-bit values)

redValue & 0xff, greenValue & 0xff, blueValue & 0xff
Accommodates the channel values packing then into one 32 bit integer using the following layout:

edited Mar 4, 2014 at 16:25

answered Mar 4, 2014 at 15:44

dinox0r

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10 Comments

Douglas Grealis Over a year ago

thanks for the reply. I initially split the pixels like so: rgbPixels[j][0] = ((rawPixels[j]>>16)&0xff); rgbPixels[j][1] = ((rawPixels[j]>>8)&0xff); rgbPixels[j][2] = (rawPixels[j]&0xff); rgbPixels[j][3] = ((rawPixels[j]>>24)&0xff); Should I leave out the alpha value?

dinox0r Over a year ago

Yes, the alpha channel should be empty (ignored, in fact) for this format.

Douglas Grealis Over a year ago

Thanks for the reply. For some reason, my resultant image is completely scrambled up. I think it could be the fact i took in the alpha values initially, but did nothing with it, and now I am using the 3 RGB values only to form the image again. What should I do with that other array set, (i'd have to create a alpha array) ? Thanks

Douglas Grealis Over a year ago

Unless I could alter the initial operations, and don't take in that alpha values. Thanks

dinox0r Over a year ago

Try ignoring the alpha and using: rgbPixels[j][0] << 16 | rgbPixels[j][1] << 8 | rgbPixels[j][2] only

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