How to convert RGB to BGR?

Question

This is probably easy, but I'm trying to convert from a source which provides colors in RGB strings to an output in BGR strings in Java. I've been busting my brain and time on shifting and Long.decode and Long.toHexString.

Feel free to also throw alpha values in there (RGBA -> ABGR), though I think I can extend the principles.

I can assume that the hex is in the form specified in the long and int decode:

0x HexDigits  
0X HexDigits 
# HexDigits

Answer 1

For 24bit colors (8 bits to each of R,G,B):

String rgbSource = getRGBSource(); //your function to get a string version of it
int in = Integer.decode(rgbSource);
int red = (in >> 16) & 0xFF;
int green = (in >> 8) & 0xFF;
int blue = (in >> 0) & 0xFF;
int out = (blue << 16) | (green << 8) | (red << 0);

Answer 2

int abgr = Integer.reverseBytes(rgba);

---

Including supporting code, with the assumption that it is safe to decide whether alpha needs adding based on the string length (consider "0xFFFFFF".length() for example):

String rgb = getRGB();

//decodes a number of hex formats and sets alpha
int rgba = rgb.length() < 8 ? 
           Long.decode(rgb + "ff").intValue() : 
           Long.decode(rgb       ).intValue();

int abgr = Integer.reverseBytes(rgba);

---

Here's a one line method:

public static String reverseRGB(String rgba) {
    return String.format("%08X",Integer.reverseBytes(Long.decode(rgba.length() < 8 ? rgba + "ff" : rgba).intValue()));
}

Answer 3

Here is how I got it to work, but I really hope there is a better way because this is awful. Please come up with a cleaner, more efficient way to do this, so I can give you rep.

long number = (rgb.length() < 8 ? Long.decode(rgb+ "ff") : Long.decode(rgb)); //decodes a number of hex formats and sets alpha
String abgrColor = (new StringBuilder())
.append((Long.toHexString((number) & 0xFF).length()==2? Long.toHexString((number) & 0xFF): "0"+Long.toHexString((number) & 0xFF)))
.append((Long.toHexString((number>>8) & 0xFF).length()==2? Long.toHexString((number>>8) & 0xFF): "0"+Long.toHexString((number>>8) & 0xFF)))
.append((Long.toHexString((number>>16) & 0xFF).length()==2? Long.toHexString((number>>16) & 0xFF): "0"+Long.toHexString((number>>16) & 0xFF)))
.append((Long.toHexString((number>>24) & 0xFF).length()==2? Long.toHexString((number>>24) & 0xFF): "0"+Long.toHexString((number>>24) & 0xFF)))
.toString();

Answer 4

If the input is a 6 character rgb string:

String bgr = rgb.substring(4,6) + rgb.substring(2,4) + rgb.substring(0,2);

If the input is an 8 character rgba string:

String abgr = rgba.substring(6,8) + rgba.substring(4,6) + rgba.substring(2,4) + rgba.substring(0,2);

If the input is an int with 8 bit channels:

String bgr = String.format( "%02X%02X%02X" , rgb & 0x00FF , (rgb>>8) & 0x00FF , (rgb>>16) & 0x00FF );
String abgr = String.format( "%02X%02X%02X%02X" , rgba & 0x00FF , (rgba>>8) & 0x00FF , (rgba>>16) & 0x00FF , (rgba>>24) & 0x00FF );
// or
String bgr = String.format( "%06X" , Integer.reverseBytes( rgb ) >> 8 );
String abgr = String.format( "%08X" , Integer.reverseBytes( rgba ) );