Max Max - 23 days ago 4
Java Question

What does this java bitshift do?

I found some bitshift code in the Java implementation of Hough Transform on Rosetta code, I understand in general what the code does, except this part:

rgbValue = (int)(((rgbValue & 0xFF0000) >>> 16) * 0.30 + ((rgbValue & 0xFF00) >>> 8) * 0.59 + (rgbValue & 0xFF) * 0.11);


I think it takes the average of all 3 pixels, that is at least what it looks like when I output the result. But how does this work? What are these magic numbers?

Method in which this function is used, pasted for convenience:


public static ArrayData getArrayDataFromImage(String filename) throws IOException
{
BufferedImage inputImage = ImageIO.read(new File(filename));
int width = inputImage.getWidth();
int height = inputImage.getHeight();
int[] rgbData = inputImage.getRGB(0, 0, width, height, null, 0, width);
ArrayData arrayData = new ArrayData(width, height);
// Flip y axis when reading image
for (int y = 0; y < height; y++)
{
for (int x = 0; x < width; x++)
{
int rgbValue = rgbData[y * width + x];

// What does this do?
rgbValue = (int)(((rgbValue & 0xFF0000) >>> 16) * 0.30 + ((rgbValue & 0xFF00) >>> 8) * 0.59 + (rgbValue & 0xFF) * 0.11);
arrayData.set(x, height - 1 - y, rgbValue);
}
}
return arrayData;
}


Answer

This is the trick that converts a 24-bit RGB value to a grayscale value using coefficients 0.3, 0.59, and 0.11 (note that these values add up to 1).

This operation (rgbValue & 0xFF0000) >>> 16 cuts out bits 17..24, and shifts them right to position 0..7, producing a value between 0 and 255, inclusive. Similarly, (rgbValue & 0xFF00) >>> 8 cuts out bits 8..16, and shifts them to position 0..7.

This Q&A talks about the coefficients, and discusses other alternatives.

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