How to convert Hex to RGB?

Question

I am trying to use this to figure out if a color is light or dark

Evaluate whether a HEX value is dark or light

Now. It takes in a int

 float calcLuminance(int rgb)
 {
      int r = (rgb & 0xff0000) >> 16;
      int g = (rgb & 0xff00) >> 8;
      int b = (rgb & 0xff);

      return (r*0.299f + g*0.587f + b*0.114f) / 256;
 }

I have a hex color though.

I tried to do this

  var color = System.Drawing.ColorTranslator.FromHtml("#FFFFFF");
  int rgb = color.R + color.G + color.B;
   var a = calcLuminance(rgb);

I got 0.11725 I thought it would have to be in the range of 0-256 or something like that.

What am I doing wrong? Do I have to covert R to an int? Or am I just way off?

Answer 1

Just convert the hex string to an integer:

int color = Convert.ToInt32("FFFFFF", 16);

Answer 2

You can use:

public string GenerateRgba(string backgroundColor, decimal backgroundOpacity)
{
 Color color = ColorTranslator.FromHtml(hexBackgroundColor);
 int r = Convert.ToInt16(color.R);
 int g = Convert.ToInt16(color.G);
 int b = Convert.ToInt16(color.B);
 return string.Format("rgba({0}, {1}, {2}, {3});", r, g, b, backgroundOpacity);
}

Link To original Post by jeremy clifton on git

Answer 3

I am trying to use this to figure out if a color is light or dark

Just use Color.GetBrightness()

---

[Edit]

I want to determine if I should use white or black for my text. So anything ≤ .5 I should use white and > .5 black?

There are a number of ways to determine what color to use on a given background, none of which are perfect.

That last link actually recommends using black/white only, but choosing a cutoff point of 0.73 instead of 0.5. I think you should just go with that, and change it if you find it doesn't work for you.

Answer 4

A little of topic, but here is an extension method to the Color struct I've created to calculate Luminance with different algorithms. Hope it helps you.

public static class ColorExtensions
{
    /// <summary>
    /// Gets the luminance of the color. A value between 0 (black) and 1 (white)
    /// </summary>
    /// <param name="color">The color.</param>
    /// <param name="algorithm">The type of luminance alg to use.</param>
    /// <returns>A value between 0 (black) and 1 (white)</returns>
    public static double GetLuminance(this Color color, LuminanceAlgorithm algorithm = LuminanceAlgorithm.Photometric)
    {
        switch (algorithm)
        {
            case LuminanceAlgorithm.CCIR601:
                return (0.2126 * color.R + 0.7152 * color.G + 0.0722 * color.B) / 255;

            case LuminanceAlgorithm.Perceived:
                return (Math.Sqrt(0.241 * Math.Pow(color.R, 2) + 0.691 * Math.Pow(color.G, 2) + 0.068 * Math.Pow(color.B, 2)) / 255);

            case LuminanceAlgorithm.Photometric:
                return (0.299 * color.R + 0.587 * color.G + 0.114 * color.B) / 255;
        }

    }

   /// <summary>
   /// The luminances
   /// </summary>
   public enum LuminanceAlgorithm
   {
       /// <summary>
       /// Photometric/digital ITU-R
       /// </summary>
       Photometric,

       /// <summary>
       /// Digital CCIR601 (gives more weight to the R and B components, as preciev by the human eye)
       /// </summary>
       CCIR601,

       /// <summary>
       /// A perceived luminance
       /// </summary>
       Perceived
   }
}

Answer 5

The problem, as I see it, is your calculation of rgb. You add the values together which gives you a number between 0 and 3*255 which clearly isn't the value your method expect. You will have to calculate it like this

int rgb = (int)color.R << 16 + (int)color.G << 8 + color.B;

which should be equivalent to this (except for the alpha-value you don't use)

int rgb = color.ToArgb();

Lastly, as you can see in Chris Haas answer, you can skip this step by converting directly to an int.

Answer 6

Your idea is OK, but your function is wrong, correct one is here:

int rgb = Convert.ToInt32("#FFFFFF", 16);
var a = calcLuminance(rgb);

float calcLuminance(int rgb)
{
    int r = (rgb & 0xff0000) >> 16;
    int g = (rgb & 0xff00) >> 8;
    int b = (rgb & 0xff);
    return (r*0.299f + g*0.587f + b*0.114f) / 256;
}

Answer 7

The ranges of the R, G and B from the Color struct are 0-255.

To get the rgb value you expect in your function, you will need to left shift accordingly:

int rgb = (int)color.R << 16 + (int)color.G << 8 + color.B;

Answer 8

calcLuminance only returns a percentage.

Answer 9

You can also convert HEX to RGB without using System.Drawing with this :

 int RGBint = Convert.ToInt32("FFD700", 16);
 byte Red = (byte)((RGBint >> 16) & 255);
 byte Green = (byte)((RGBint >> 8) & 255);
 byte Blue = (byte)(RGBint & 255);
 //Color.FromRgb(Red, Green, Blue);