How to Convert Rgb Color to Int in Java

Convert RGB values to Integer

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;

Convert integer color value to RGB

Use this

String hexColor = String.format("#%06X", (0xFFFFFF & intColor));

We know lenght of color value in HEX is 6. So you see 6 here. %06X matches the result coming from (0xFFFFFF & intColor) and if length is less than 6, it makes result with 6 by appending ZERO to left side of result. And you see #, so this # char gets appended to result and finally you get a HEX COLOR value.

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Update since API 26

Since API 26, new methods Color.valueOf(....) has been introduced to convert colors for similar reason. you can use it like

// sRGB
Color opaqueRed = Color.valueOf(0xffff0000); // from a color int
Color translucentRed = Color.valueOf(1.0f, 0.0f, 0.0f, 0.5f);

// Wide gamut color
ColorSpace sRgb = ColorSpace.get(ColorSpace.Named.SRGB);
@ColorLong long p3 = Color.pack(1.0f, 1.0f, 0.0f, 1.0f, sRgb);
Color opaqueYellow = Color.valueOf(p3); // from a color long

// CIE L*a*b* color space
ColorSpace lab = ColorSpace.get(Named.CIE_LAB);
Color green = Color.valueOf(100.0f, -128.0f, 128.0f, 1.0f, lab);

mView.setBackgroundColor(opaqueRed.toArgb());
mView2.setBackgroundColor(green.toArgb());
mView3.setBackgroundColor(translucentRed.toArgb());
mView4.setBackgroundColor(opaqueYellow.toArgb());

Java Color encode RGB into single integer?

The Javadocs for Color explain that the Java Color class contains an alpha channel as well as RGB--so a color that started with all zeroes would be completely transparent. The color represented by 0xff000000 is opaque black.

In this case, of course, the author has made an error. The single-argument constructor ignores the high byte. Instead, it should have used new Color(0xffc00000, true) if he wanted to specify the alpha (or, more simply, new Color(0xc0, 0, 0, 0xff)).

How to convert getRGB(x,y) integer pixel to Color(r,g,b,a) in Java?

If I'm guessing right, what you get back is an unsigned integer of the form 0xAARRGGBB, so

int b = (argb)&0xFF;
int g = (argb>>8)&0xFF;
int r = (argb>>16)&0xFF;
int a = (argb>>24)&0xFF;

would extract the color components. However, a quick look at the docs says that you can just do

Color c = new Color(argb);

or

Color c = new Color(argb, true);

if you want the alpha component in the Color as well.

UPDATE

Red and Blue components are inverted in original answer, so the right answer will be:

int r = (argb>>16)&0xFF;
int g = (argb>>8)&0xFF;
int b = (argb>>0)&0xFF;

updated also in the first piece of code

Convert color RGB to HEX (int result)

Suppose you string is

String col = "0xF82619";

try:

view.setBackgroundColor(Color.parseColor(col));

Convert RGB values to binary to decimal

You need the conversion from double because you cannot store the double value into 8 bit.

Your double values are between 0.0 and 1.0. You can see them as the proportion of color used (e.g. 0.33333333 in yellow means that one-third of the possible yellow is used). As you can see such a double can have many decimal places, which means we need a lot of memory (64-bit) to store such a color.

Your function now tries to store the double value into only 8 bit (256 values). As we said the double value can be seen as a portion (between 0 and 1) and the function calculates the same for 8 bit (between 0 and 255). This can simply be done by multiplying the double value with 255. For the example with yellow (0.33333333 of yellow is used) it is: 0.33333333 * 255 = 84,99999915. The meaning is still the same 84,99999915 yellow parts of 255 yellow parts are used, which is still a third.
In order to get a compressed version of this number, it is rounded to the next integer value. In our example, this is 85, which is really close to the actual portion, but we save a lot of memory.

It makes also sense for the lowest double value 0.0, which is converted to the lowest int value 0. The highest double value 1.0 is converted to 255 (highest 8-bit number).

In conclusion, we convert a double (64-bit) into an only 8-bit number, which has the same proportion of the color, but it is not as accurate.

Edit: As there is also a confusion with the 255: 8 bit can store 256 values (0 to 255). If you can choose 256 as a color value somewhere they use the range 1-256 without 0. Essentially it is the same one shifted by 1.

How to convert RGB color to grayscale in Java

The float components are in the range of [0.0, 1.0]. You are adding the scaled RGB components together and then casting them to int:

rgb[0] = rgb[1] = rgb[2] = (int)(0.299 * rgb[0] + 0.587 * rgb[1] + 0.114 * rgb[2]);
                            ^^^ Cast to int

You would get the color black if all of the values in the rgb array are zero. This is most likely because the cast to int truncates the resulting floating-point values down to zero. Since rgb is an array of floats, you should cast result to float.

rgb[0] = rgb[1] = rgb[2] =  (float)(0.299 * rgb[0] + 0.587 * rgb[1] + 0.114 * rgb[2]);
                           ^ Key change, cast to float

How do i convert RGB color to hexadecimal in Java

public static int colorToInt(int r, int g, int b) {
    int result = 0;
    result += r;
    result = result << 8;
    result += g;
    result = result << 8;
    result += b;

    return result;
}

Here's an example:

System.out.println(0xffffff); // 16777215
System.out.println(colorToInt(255, 255, 255)); // 16777215

Repl: https://repl.it/repls/IrritatingTwinConsulting.

This function accepts int, instead of double, but I don't think it's a problem. Hope I understood your question correctly.

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