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Home/ Questions/Q 6936627
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T12:15:29+00:00 2026-05-27T12:15:29+00:00

I’m trying to write a function to shift the hue of an RGB color.

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I’m trying to write a function to shift the hue of an RGB color. Specifically I’m using it in an iOS app, but the math is universal.

The graph below shows how the R, G, and B values change with respect to the hue.

Graph of RGB values across hues

Looking at that it seems like it should be a relatively simple to write a function to shift the hue without doing any nasty conversions to a different color format which would introduce more error (which could be an issue if continue applying small shifts to a color), and I suspect would be more computationally expensive.

Here is what I have so far which sort of works. It works perfectly if you’re shifting from pure yellow or cyan or magenta but otherwise it gets a little squiffy in some places.

Color4f ShiftHue(Color4f c, float d) {
    if (d==0) {
        return c;
    }
    while (d<0) {
        d+=1;
    }

    d *= 3;

    float original[] = {c.red, c.green, c.blue};
    float returned[] = {c.red, c.green, c.blue};

    // big shifts
    for (int i=0; i<3; i++) {
        returned[i] = original[(i+((int) d))%3];
    }
    d -= (float) ((int) d);
    original[0] = returned[0];
    original[1] = returned[1];
    original[2] = returned[2];

    float lower = MIN(MIN(c.red, c.green), c.blue);
    float upper = MAX(MAX(c.red, c.green), c.blue);

    float spread = upper - lower;
    float shift  = spread * d * 2;

    // little shift
    for (int i = 0; i < 3; ++i) {
        // if middle value
        if (original[(i+2)%3]==upper && original[(i+1)%3]==lower) {
            returned[i] -= shift;
            if (returned[i]<lower) {
                returned[(i+1)%3] += lower - returned[i];
                returned[i]=lower;
            } else
                if (returned[i]>upper) {
                    returned[(i+2)%3] -= returned[i] - upper;
                    returned[i]=upper;
                }
            break;
        }
    }

    return Color4fMake(returned[0], returned[1], returned[2], c.alpha);
}

I know you can do this with UIColors and shift the hue with something like this:

CGFloat hue;
CGFloat sat;
CGFloat bri;
[[UIColor colorWithRed:parent.color.red green:parent.color.green blue:parent.color.blue alpha:1] getHue:&hue saturation:&sat brightness:&bri alpha:nil];
hue -= .03;
if (hue<0) {
    hue+=1;
}
UIColor *tempColor = [UIColor colorWithHue:hue saturation:sat brightness:bri alpha:1];
const float* components= CGColorGetComponents(tempColor.CGColor);
color = Color4fMake(components[0], components[1], components[2], 1);

but I’m not crazy about that as It only works in iOS 5, and between allocating a number of color objects and converting from RGB to HSB and then back it seems pretty overkill.

I might end up using a lookup table or pre-calculate the colors in my application, but I’m really curious if there’s a way to make my code work. Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T12:15:30+00:00Added an answer on May 27, 2026 at 12:15 pm

    Edit per comment changed “are all” to “can be linearly approximated by”.
    Edit 2 adding offsets.


    Essentially, the steps you want are

    RBG->HSV->Update hue->RGB
    

    Since these can be approximated by linear matrix transforms (i.e. they are associative), you can perform it in a single step without any nasty conversion or loss of precision. You just multiple the transform matrices with each other, and use that to transform your colors.

    There’s a quick step by step here http://beesbuzz.biz/code/hsv_color_transforms.php

    Here’s the C++ code (With the saturation and value transforms removed):

    Color TransformH(
        const Color &in,  // color to transform
        float H
    )
    {
      float U = cos(H*M_PI/180);
      float W = sin(H*M_PI/180);
    
      Color ret;
      ret.r = (.299+.701*U+.168*W)*in.r
        + (.587-.587*U+.330*W)*in.g
        + (.114-.114*U-.497*W)*in.b;
      ret.g = (.299-.299*U-.328*W)*in.r
        + (.587+.413*U+.035*W)*in.g
        + (.114-.114*U+.292*W)*in.b;
      ret.b = (.299-.3*U+1.25*W)*in.r
        + (.587-.588*U-1.05*W)*in.g
        + (.114+.886*U-.203*W)*in.b;
      return ret;
    }
    
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