Brad
Brad

Reputation: 21160

How do I adjust the brightness of a color?

I would like to darken an existing color for use in a gradient brush. Could somebody tell me how to do this please?

C#, .net 2.0, GDI+

Upvotes: 31

Views: 58552

Answers (7)

torpid prey
torpid prey

Reputation: 242

Similar to accepted answer but this uses a percentage of the difference between each rgb value and 255 to determine each new rgb value.

VB.NET

Function Lighten(orig As Color, Optional percent As Integer = 80) As Color
    'get remainders
    Dim rr As Integer = 255 - orig.R
    Dim gr As Integer = 255 - orig.G
    Dim br As Integer = 255 - orig.B

    'add a percentage of the remainder, plus original value
    Dim r As Integer = CInt(percent / 100 * rr) + orig.R
    Dim g As Integer = CInt(percent / 100 * gr) + orig.G
    Dim b As Integer = CInt(percent / 100 * br) + orig.B

    Return Color.FromArgb(r, g, b)
End Function

Function Darken(orig As Color, Optional percent As Integer = 80) As Color
    'subtract the percentage of the original value from the original value
    Dim r As Integer = orig.R - CInt(percent / 100 * orig.R)
    Dim g As Integer = orig.G - CInt(percent / 100 * orig.G)
    Dim b As Integer = orig.B - CInt(percent / 100 * orig.B)

    Return Color.FromArgb(r, g, b)
End Function

C#

public Color Lighten(Color orig, int percent = 80)
{
    // get remainders
    int rr = 255 - orig.R;
    int gr = 255 - orig.G;
    int br = 255 - orig.B;

    // add a percentage of the remainder, plus original value
    int r = System.Convert.ToInt32(percent / (double)100 * rr) + orig.R;
    int g = System.Convert.ToInt32(percent / (double)100 * gr) + orig.G;
    int b = System.Convert.ToInt32(percent / (double)100 * br) + orig.B;

    return Color.FromArgb(r, g, b);
}

public Color Darken(Color orig, int percent = 80)
{
    // subtract the percentage of the original value from the original value
    int r = orig.R - System.Convert.ToInt32(percent / (double)100 * orig.R);
    int g = orig.G - System.Convert.ToInt32(percent / (double)100 * orig.G);
    int b = orig.B - System.Convert.ToInt32(percent / (double)100 * orig.B);

    return Color.FromArgb(r, g, b);
}

Upvotes: 0

dommer
dommer

Reputation: 19810

Here's some C# code for the conversions Richard mentioned:

RGB to HSL / HSL to RGB in C#

Upvotes: 11

VB.net Coder
VB.net Coder

Reputation: 17

You must keep track that the value does not extend below 0 or above 255

Best approach is to use Math.Max/Math.MIn

dim newValue as integer = ...
'correct value if it is below 0 or above 255
newValue = Math.Max(Math.Min(newValue,255),0)

Upvotes: 2

TugboatCaptain
TugboatCaptain

Reputation: 4328

While the aforementioned methods do darken the color but they adjust the hue way to much so the result doesn't look very good. The best answer is to use Rich Newman's HSLColor class and adjust the luminosity.

public Color Darken(Color color, double darkenAmount) {
    HSLColor hslColor = new HSLColor(color);
    hslColor.Luminosity *= darkenAmount; // 0 to 1
    return hslColor;
}

Upvotes: 5

Richard
Richard

Reputation: 109005

Convert from RGB to HSV (or HSL), then adjust the V (or L) down and then convert back.

While System.Drawing.Color provides methods to get hue (H), saturation (S) and brightness it does not provide much in the way of other conversions, notable nothing to create a new instance from HSV (or HSV values), but the conversion is pretty simple to implement. The wikipedia articles give decent converage, starting here: "HSL and HSV".

Upvotes: 19

Alex
Alex

Reputation: 611

You could also try using

ControlPaint.Light(baseColor, percOfLightLight)

ControlPaint.Light

or

ControlPaint.Dark(baseColor, percOfDarkDark)

ControlPaint.Dark

Upvotes: 32

Marc Gravell
Marc Gravell

Reputation: 1062780

As a simple approach, you can just factor the RGB values:

    Color c1 = Color.Red;
    Color c2 = Color.FromArgb(c1.A,
        (int)(c1.R * 0.8), (int)(c1.G * 0.8), (int)(c1.B * 0.8));

(which should darken it; or, for example, * 1.25 to brighten it)

Upvotes: 41

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