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
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.