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Color code to int (NumberFormatExeption) [duplicate]

So in a BufferedImage, you receive a single integer that has the RGB values represented in it. So far I use the following to get the RGB values from it:
// rgbs is an array of integers, every single integer represents the
// RGB values combined in some way
int r = (int) ((Math.pow(256,3) + rgbs[k]) / 65536);
int g = (int) (((Math.pow(256,3) + rgbs[k]) / 256 ) % 256 );
int b = (int) ((Math.pow(256,3) + rgbs[k]) % 256);
And so far, it works.
What I need to do is figure out how to get an integer so I can use BufferedImage.setRGB(), because that takes the same type of data it gave me.

I think the code is something like:
int rgb = red;
rgb = (rgb << 8) + green;
rgb = (rgb << 8) + blue;
Also, I believe you can get the individual values using:
int red = (rgb >> 16) & 0xFF;
int green = (rgb >> 8) & 0xFF;
int blue = rgb & 0xFF;

int rgb = ((r&0x0ff)<<16)|((g&0x0ff)<<8)|(b&0x0ff);
If you know that your r, g, and b values are never > 255 or < 0 you don't need the &0x0ff
Additionaly
int red = (rgb>>16)&0x0ff;
int green=(rgb>>8) &0x0ff;
int blue= (rgb) &0x0ff;
No need for multipling.

if r, g, b = 3 integer values from 0 to 255 for each color
then
rgb = 65536 * r + 256 * g + b;
the single rgb value is the composite value of r,g,b combined for a total of 16777216 possible shades.

int rgb = new Color(r, g, b).getRGB();

To get individual colour values you can use Color like following for pixel(x,y).
import java.awt.Color;
import java.awt.image.BufferedImage;
Color c = new Color(buffOriginalImage.getRGB(x,y));
int red = c.getRed();
int green = c.getGreen();
int blue = c.getBlue();
The above will give you the integer values of Red, Green and Blue in range of 0 to 255.
To set the values from RGB you can do so by:
Color myColour = new Color(red, green, blue);
int rgb = myColour.getRGB();
//Change the pixel at (x,y) ti rgb value
image.setRGB(x, y, rgb);
Please be advised that the above changes the value of a single pixel. So if you need to change the value entire image you may need to iterate over the image using two for loops.

Color matching library

I'm looking for a way to get a matching color(s) for any given color.
Is there a way to do this either programmatically or using a library (in Java/Android)?
Thanks!
(As in a color that goes well with said color, This is a good example: colorcombos.com/colors/FF0000)

A simple method for getting the most contrasting color, would be to bounce each RGB color as far as possible. By this I mean a value of 00-7F becomes FF, and a value of 80-FF becomes 00. The result will always be saturated.
If you want the complimentary color to stay consistent with the original color, e.g a pastel red becomes a pastel cyan, things get more complicated, but it all depends on the color model you use, e.g. RGB vs HSV vs others.
You could use the formula of 0xFF - value for each RGB color, but a gray will stay gray, so that's not always good.
As for how to work with the RGB color values, here's how to find most contrasting color of a color given as the hex string.
String color = "2E8B57" // SeaGreen
int r1 = Integer.parseInt(color.substring(0, 2), 16);
int g1 = Integer.parseInt(color.substring(2, 2), 16);
int b1 = Integer.parseInt(color.substring(4, 2), 16);
int r2 = (r1 < 0x80 ? 0xFF : 0x00);
int g2 = (g1 < 0x80 ? 0xFF : 0x00);
int b2 = (b1 < 0x80 ? 0xFF : 0x00);
String newColor = String.format("%02x%02x%02x", r2, g2, b2);
// newColor = "FF00FF" (Fuchsia)
Using the other way, you'd use:
int r2 = 0xFF - r1;
int g2 = 0xFF - g1;
int b2 = 0xFF - b1;
String newColor = String.format("%02x%02x%02x", r2, g2, b2);
// newColor = "D174A8" (Hopbush)
Color names are from www.htmlcsscolor.com: SeaGreen, Fuchsia, Hopbush

Convert negative rgb value to int [duplicate]

Ok so I'm working on a program that takes in an image, isolates a block of pixels into an array, and then gets each individual rgb value for each pixel in that array.
When I do this
//first pic of image
//just a test
int pix = myImage.getRGB(0,0)
System.out.println(pix);
It spits out -16106634
I need to get the (R, G, B) value out of this int value
Is there a formula, alg, method?

The BufferedImage.getRGB(int x, int y) method always returns a pixel in the TYPE_INT_ARGB color model. So you just need to isolate the right bits for each color, like this:
int pix = myImage.getRGB(0, 0);
int r = (pix >> 16) & 0xFF;
int g = (pix >> 8) & 0xFF;
int b = pix & 0xFF;
If you happen to want the alpha component:
int a = (pix >> 24) & 0xFF;
Alternatively you can use the Color(int rgba, boolean hasalpha) constructor for convenience (at the cost of performance).

int pix = myImage.getRGB(0,0);
Color c = new Color(pix,true); // true for hasalpha
int red = c.getRed();
int green = c.getGreen();
int blue = c.getBlue();

signed int to unsigned byte and vice versa

I have a signed value e.g. -7368817
when I cast it to byte it will be something like: -113
I convert it to unsigned byte bye & 0xff and it will be something like 143
now I manipulate this byte value and after that I want to whole way back! to get the signed integer of the new byte. :)
Update
The whole story:
I have an image which is 8-bit depth and gray scale! it means that all pixels are presented using 1 byte
BufferedImage image = ImageIO.read(new File("my_grayscal_8bit_photo.jpg"));
int intPixel = image.getRGB(1, 1);
now I need some bit manipulation in this pixel but since it is int I must convert it to byte first:
byte bytePixel = (byte) intPixel;
and to make it unsigned:
int intPixel2 = bytePixel & 0xff;
now I do my bit manipulation and want to convert it to int and do:
image.setRGB(1, 1, neworiginalint);

The getRGB(int x, int y) method always returns an int pixel in the TYPE_INT_ARGB color model. To manually extract the red, green, blue and alpha values for the pixel you can do this:
int pixel = image.getRGB(1, 1);
int a = (pixel >> 24) & 0xFF;
int r = (pixel >> 16) & 0xFF;
int g = (pixel >> 8) & 0xFF;
int b = pixel & 0xFF;
Or use the Color(int rgba, boolean hasalpha) constructor for convenience (at the cost of performance). Once you've manipulated the red, green, blue and alpha values (in the range of 0 to 255) you can recombine them back into an int for setting pixels:
int newPixel = (a << 24) | (r << 16) | (g << 8) | b;
Using the -7368817 pixel you mentioned with this code, the alpha is 255 (so no transparency) and the red, green and blue values are all 143. Since you're dealing with grayscale you could just pick any of red, green or blue to get the gray value. However on setting the pixel you're going to have set all three to maintain the grayscale since it's RGB. You could shortcut it a bit like so:
int pixel = image.getRGB(1, 1);
// extract your gray value from blue, assume red and green are same
int gray = pixel & 0xFF;
// this method does your manipulation on the gray value, 0 to 255
gray = manipulate(gray);
// recombine back into int, preserving the original alpha
int newPixel = (pixel & 0xFF000000) | (gray << 16) | (gray << 8) | gray;
// now you can set your new pixel
image.setRGB(1, 1, nexPixel);
Basically the trick is to use int as your unsigned byte. Just make sure you keep the values from 0 to 255 and everything should work fine.

Are you asking how to manipulate the least-significant byte in an signed int, without touching the more significant byte?
This can be done like this:
int i = -7368817; // 0xFF8F 8F8F
int b = i & 0xFF; // "signed byte", = 0x8F = 143
b += 0x2C; // your manipulation, result = 0xBB
i = (i & 0xFFFFFF00) | (b & 0xFF); // result of LSB modification = 0xFF8F8FBB
or in one step: i = (i & 0xFFFFFF00) | ((i + 0x2C) & 0xFF); if it is a simple manipulation.
If the manipulation can't ever produce an overflow, you can simply do it on the whole int:
i ^= 0x34; // i = 0xFF8F8FBB

Convert RGB values to Integer

So in a BufferedImage, you receive a single integer that has the RGB values represented in it. So far I use the following to get the RGB values from it:
// rgbs is an array of integers, every single integer represents the
// RGB values combined in some way
int r = (int) ((Math.pow(256,3) + rgbs[k]) / 65536);
int g = (int) (((Math.pow(256,3) + rgbs[k]) / 256 ) % 256 );
int b = (int) ((Math.pow(256,3) + rgbs[k]) % 256);
And so far, it works.
What I need to do is figure out how to get an integer so I can use BufferedImage.setRGB(), because that takes the same type of data it gave me.

I think the code is something like:
int rgb = red;
rgb = (rgb << 8) + green;
rgb = (rgb << 8) + blue;
Also, I believe you can get the individual values using:
int red = (rgb >> 16) & 0xFF;
int green = (rgb >> 8) & 0xFF;
int blue = rgb & 0xFF;

int rgb = ((r&0x0ff)<<16)|((g&0x0ff)<<8)|(b&0x0ff);
If you know that your r, g, and b values are never > 255 or < 0 you don't need the &0x0ff
Additionaly
int red = (rgb>>16)&0x0ff;
int green=(rgb>>8) &0x0ff;
int blue= (rgb) &0x0ff;
No need for multipling.

if r, g, b = 3 integer values from 0 to 255 for each color
then
rgb = 65536 * r + 256 * g + b;
the single rgb value is the composite value of r,g,b combined for a total of 16777216 possible shades.

int rgb = new Color(r, g, b).getRGB();

To get individual colour values you can use Color like following for pixel(x,y).
import java.awt.Color;
import java.awt.image.BufferedImage;
Color c = new Color(buffOriginalImage.getRGB(x,y));
int red = c.getRed();
int green = c.getGreen();
int blue = c.getBlue();
The above will give you the integer values of Red, Green and Blue in range of 0 to 255.
To set the values from RGB you can do so by:
Color myColour = new Color(red, green, blue);
int rgb = myColour.getRGB();
//Change the pixel at (x,y) ti rgb value
image.setRGB(x, y, rgb);
Please be advised that the above changes the value of a single pixel. So if you need to change the value entire image you may need to iterate over the image using two for loops.