Int argb color output strange value

Question

I'm trying to create small app where is using random colors.

Random rnd = new Random();
        int color1 = Color.argb(255, rnd.nextInt(256), rnd.nextInt(256), rnd.nextInt(256));
        int color2 = Color.argb(255, rnd.nextInt(256), rnd.nextInt(256), rnd.nextInt(256));
        int color3 = Color.argb(255, rnd.nextInt(256), rnd.nextInt(256), rnd.nextInt(256));

But in color1, color2 and color3 saved values such as "-11338194". Is it possible to take argb value? (Such as "255255255255" or something) Thank you!

Think about what 32 bit integer is represented by -11338194... then work out what its 4 8-bit values are... — Jon Skeet, Jul 04 '16 at 18:58
@JonSkeet , ghm. Byte b = (byte) color1 doesn't work. Sorry, I'm just learning to code — Dmitry Ghraz, Jul 04 '16 at 19:05
Well you'd have to define "doesn't work" for that to make sense... it's not clear what you expected. — Jon Skeet, Jul 04 '16 at 19:06
@JonSkeet , when I'm trying to convert into byte is has this value: (color1 = -9003271 and byte value is -7) — Dmitry Ghraz, Jul 04 '16 at 19:09
Right, but it's not clear what you'd expected it to be or why. — Jon Skeet, Jul 04 '16 at 19:09
@JonSkeet , i'm trying to get argb value such as "255 255 255 255" or equals — Dmitry Ghraz, Jul 04 '16 at 19:10
http://stackoverflow.com/a/18037185/6525469 this is how rgb values are stored in an int. You may revert using http://stackoverflow.com/a/28111503/6525469 — Sakchham, Jul 04 '16 at 19:12

score 1 · Answer 1 · edited May 23 '17 at 12:16

1

Try this code,

Random rnd = new Random();
        int color1 = Color.argb(255, rnd.nextInt(256 - 0), rnd.nextInt(256 - 0), rnd.nextInt(256 - 0));
        int color2 = Color.argb(255, rnd.nextInt(256 - 0), rnd.nextInt(256 - 0), rnd.nextInt(256 - 0));
        int color3 = Color.argb(255, rnd.nextInt(256 - 0), rnd.nextInt(256 - 0), rnd.nextInt(256 - 0));

Reference for Color.argb()

Generate Random number between range

edited May 23 '17 at 12:16

Community

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answered Jul 04 '16 at 19:05

madhan kumar

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Thanks! But it don't change anything :( – Dmitry Ghraz Jul 04 '16 at 19:09
@Dimitry Still it generating negative values? – madhan kumar Jul 04 '16 at 19:11
yes, it is still generating negative strange values – Dmitry Ghraz Jul 04 '16 at 19:12

score 1 · Answer 2 · answered Oct 19 '16 at 16:22

A Java color is represented by a 32 bit integer in ARGB format.

That means the highest 8 bit is an alpha value, 255 means full opacity while 0 means transparency. You generate colors with alpha value of 255.

An Integer is a signed number, its most significant bit tells if it is negative. As you set all the first 8 bits to 1, effectively all Colors will be negative numbers if you print it to the screen.

Example:

 System.err.println("Color="+new java.awt.Color(0,0,255,0).getRGB());
 gives 255 as you expected - note that this is a fully transparent blue

 System.err.println("Color="+java.awt.Color.RED.getRGB());
 gives -65536, as the alpha channel value is 255 making the int negative.

If you only want to see the RGB values, simply do a logical AND to truncate the alpha channel bits which make the decimal numeric representation negative:

 System.err.println("Color="+(java.awt.Color.RED.getRGB() & 0xffffff));
 gives you 16711680

Alternatively you can get the representation of the color in hex as:

System.err.println("Color="+String.format("%X",java.awt.Color.RED.getRGB() & 0xffffff));
which gives FF0000

(This example is from desktop JavaSE, you may find subtle differences in Android.) — Gee Bee, Oct 19 '16 at 16:23

Int argb color output strange value

2 Answers2