What does "& 0xff" do?

Question

I am trying to understand the code below where b is a given integer and image is an image.

I understand that if the RGB value at given point i,j is greater than b then set that pixel to white else set to black. so would convert the image into black and white.

However I am lost to what (& 0xff) actually does, I am guessing its a kind of binary shift?

if ((image.getRGB(i, j) & 0xff) > b) {     image.setRGB(i, j, 0xffffff) ; } else {     image.setRGB(i, j, 0x000000); } 

Answer 1

It's a so-called mask. The thing is, you get the RGB value all in one integer, with one byte for each component. Something like 0xAARRGGBB (alpha, red, green, blue). By performing a bitwise-and with 0xFF, you keep just the last part, which is blue. For other channels, you'd use:

int alpha = (rgb >>> 24) & 0xFF; int red   = (rgb >>> 16) & 0xFF; int green = (rgb >>>  8) & 0xFF; int blue  = (rgb >>>  0) & 0xFF; 

In the alpha case, you can skip & 0xFF, because it doesn't do anything; same for shifting by 0 in the blue case.

Answer 2

The

& 0xFF 

is getting one of the color components (either red or blue, I forget which).

If the color mask is not performed, consider RGB (0, 127, 0), and the threshold 63. The getRGB(...) call would return

(0 * 256 * 256) + (127 * 256) + 0 = 32512 

Which is clearly more than the threshold 63. But the intent was to ignore the other two color channels. The bitmask gets only the lowest 8 bits, with is zero.

The

> b 

is checking if the color is brighter than a particular threshold, 'b'.

If the threshold is exceeded, the pixel is colored white, using

image.setRGB(i,j,0xffffff) 

... otherwise it is colored black, using

image.setRGB(i,j,0x000000) 

So it is a conversion to black and white based on a simple pixel-by-pixel threshold on a single color channel.