Converting color value from float 0..1 to byte 0..255

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What would be the correct way of converting color value from float to byte? At first I thought b=f*255.0 should do it, but now I'm thinking, that in this case only the exact 1.0 will be converted to 255, but 0.9999 will already be 254 which is probably not what I want...

It seems that b=f*256.0 would be better except that it would have an unwanted case of making 256 in the case of exact 1.0.

In the end I'm using this:

#define F2B(f) ((f) >= 1.0 ? 255 : (int)((f)*256.0))

1.0 is the only case that can go wrong, so handle that case separately:

b = floor(f >= 1.0 ? 255 : f * 256.0)

Also, it might be worth forcing that f really is 0<=f<=1 to avoid incorrect behaviour due to rounding errors (eg. f=1.0000001).

f2 = max(0.0, min(1.0, f))
b = floor(f2 == 1.0 ? 255 : f2 * 256.0)

Alternative safe solutions:

b = (f >= 1.0 ? 255 : (f <= 0.0 ? 0 : (int)floor(f * 256.0)))


b = max(0, min(255, (int)floor(f * 256.0)))

Float to Byte for Color32, Question by nyonge � May 30, 2017 at 10:20 PM � coloralphafloatbytecolor32. Float to Byte for Color32. How do I take a float (0f to 1f) and convert that to a byte ( not a byte array) for a Color32's alpha channel (0-255)? modulo % 256 is not strictly necessary, since only the right-most byte of the int value you pass is used. I have implement a function that converts a float color channel (R, G, B) to a byte value. E.g.: 0.0f -> 0 1.0f -> 255 unsigned char RGBImage::convertColorChannel(float f) { } How can I get

I've always done round(f * 255.0).

There is no need for the testing (special case for 1) and/or clamping in other answers. Whether this is a desirable answer for your purposes depends on whether your goal is to match input values as closely as possible [my formula], or to divide each component into 256 equal intervals [other formulas].

The possible downside of my formula is that the 0 and 255 intervals only have half the width of the other intervals. Over years of usage, I have yet to see any visual evidence that that is bad. On the contrary, I've found it preferable to not hit either extreme until the input is quite close to it - but that is a matter of taste.

The possible upside is that [I believe] the relative values of R-G-B components are (slightly) more accurate, for a wider range of input values. Though I haven't tried to prove this, that is my intuitive sense, given that for each component I round to get the closest available integer. (E.g. I believe that if a color has G ~= 2 x R, this formula will more often stay close to that ratio; though the difference is quite small, and there are many other colors that the 256 formula does better on. So it may be a wash.)

In practice, either 256 or 255-based approaches seem to provide good results.

Another way to evaluate 255 vs 256, is to examine the other direction - converting from 0..255 byte to 0.0..1.0 float.

The formula that converts 0..255 integer values to equally spaced values in range 0.0..1.0 is:

f = b / 255.0

Going in this direction, there is no question as to whether to use 255 or 256: the above formula is the formula that yields equally spaced results. Observe that it uses 255.

To understand the relationship between the 255 formulas in the two directions, consider this diagram, if you only had 2 bits, hence values integer values 0..3:

Diagram using 3 for two bits, analogous to 255 for 8 bits. Conversion can be from top to bottom, or from bottom to top:

0 --|-- 1 --|-- 2 --|-- 3  
0 --|--1/3--|--2/3--|-- 0
   1/6     1/2     5/6

The | are the boundaries between the 4 ranges. Observe that in the interior, the float values and the integer values are at the midpoints of their ranges. Observe that the spacing between all values is constant in both representations.

If you grasp these diagrams, you will understand why I favor 255-based formulas over 256-based formulas.

Claim: If you use / 255.0 when going from byte to float, but you don't use round(f * 255.0) when going to byte from float, then the "average round-trip" error is increased. Details follow.

This is most easily measured by starting from float, going to byte, then back to float. For a simple analysis, use the 2-bit "0..3" diagrams.

Start with a large number of float values, evenly spaced from 0.0 to 1.0. THe round-trip will group all these values at the 4 values. The diagram has 6 half-interval-length ranges: 0..1/6, 1/6..1/3, .., 5/6..1 For each range, the average round-trip error is half the range, so 1/12 (Minimum error is zero, maximum error is 1/6, evenly distributed). All the ranges give that same error; 1/12 is the overall average error when round trip.

If you instead use any of the * 256 or * 255.999 formulas, most of the round-trip results are the same, but a few are moved to the adjacent range. Any change to another range increases the error; for example if the error for a single float input previously was slightly less than 1/6, returning the center of an adjacent range results in an error slightly more than 1/6. E.g. 0.18 in optimal formula => byte 1 => float 1/3 ~= 0.333, for error |0.33-0.18| = 0.147; using a 256 formula => byte 0 => float 0 , for error 0.18, which is an increase from the optimal error 0.147.

Diagrams using * 4 with / 3. Conversion is from one line to the next. Notice the uneven spacing of the first line: 0..3/8, 3/8..5/8, 5/8..1. Those distances are 3/8, 2/8, 3/8. Notice the interval boundaries of last line are different than first line.

         1/4     1/2     3/4
=> 0------|-- 1 --|-- 2 --|------3  

=> 0----|---1/3---|---2/3---|----0
       1/6       1/2       5/6

The only way to avoid this increased error, is to use some different formula when going from byte to float. If you strongly believe in one of the 256 formulas, then I'll leave it to you to determine the optimal inverse formula. (Per byte value, it should return the midpoint of the float values which became that byte value. Except 0 to 0, and 3 to 1. Or perhaps 0 to 1/8, 3 to 7/8! In the diagram above, it should take you from middle line back to top line.)

But now you will have the difficult-to-defend situation that you have taken equally-spaced byte values, and converted them to non-equally-spaced float values.

Those are your options if you use any value other than exactly 255, for integers 0..255: Either an increase in average round-trip error, or non-uniformly-spaced values in the float domain.

How do you convert an int byte colour value (0-255) into a float , How do you convert an int byte colour value (0-255) into a float / double value (0- 1) in C#? - c#. In WxWidget colors are represented by a RGB integer triplet. To interact with other libraries using a [0.0-1.0] float triplet representation, a conversion is needed. Is there such a conversion fun

Why not try something like


Gets rid of the special case f==1 but 0.999 is still 255

colors — Matplotlib 1.2.1 documentation, A module for converting numbers or color arguments to RGB or RGBA. RGB and RGBA are sequences of, respectively, 3 or 4 floats in the range 0-1. Out-of- range values are mapped to -1 if low and ncolors if high; these are converted to Float dtypes are preserved; integer types with two bytes or smaller are converted to� "Converting" a float in the range 0.0 to 1.0 to an int results in values of either 0 or 1. Converting by multiplying by 255, then truncating would result in values from 0 to 255. thescreamingdrills

The accepted solution failed when it compare float as it was integer.

This code work just fine:

float f;
uint8_t i;
//byte to float
f =CLAMP(((float)((i &0x0000ff))) /255.0, 0.0, 1.0);
//float to byte
i =((uint8_t)(255.0f *CLAMP(f, 0.0, 1.0)));

if you don't have CLAMP:

#define CLAMP(value, min, max) (((value) >(max)) ? (max) : (((value) <(min)) ? (min) : (value)))

Or for full RGB:

integer_color =((uint8_t)(255.0f *CLAMP(float_color.r, 0.0, 1.0)) <<16) |
               ((uint8_t)(255.0f *CLAMP(float_color.g, 0.0, 1.0)) <<8) |
               ((uint8_t)(255.0f *CLAMP(float_color.b, 0.0, 1.0))) & 0xffffff;

float_color.r =CLAMP(((float)((integer_color &0xff0000) >>16)) /255.0, 0.0, 1.0);
float_color.g =CLAMP(((float)((integer_color &0x00ff00) >>8)) /255.0, 0.0, 1.0);
float_color.b =CLAMP(((float)((integer_color &0x0000ff))) /255.0, 0.0, 1.0);

Color, Boolean � Byte � Character � Character.Subset � Character. Added in API level 1 The color long encoding relies on half-precision float values (fp16). If you simply need to convert a color int into a color long, use pack(int) . These gradients show how the RGB color 0, 1, 255 changes by changing the brightness by 10 percent. The first figure shows a shift by +10% for each color and the second figure -10%. Similar to the brightness gradients but the following saturation gradients show a change of the RGB color 0, 1, 255 by changing the saturation by 10% instead.

What do you mean by correct way of converting a color value from float to byte? Do you mean that if you choose uniform random real numbers from the range [0,1[ that they will uniquely distributed among the 256 bins from 0 to 255?

To make things easier we assume that instead of a float value we have a real number and instead of int we want to convert to a two bit integer, something like a uint_2 - a integer number representation that consists of exactly two bits. This would mean that our unit2_t can have the values 00b, 01b, 10b and 11b (the b denotes that we have here a binary number. This is also known as Intel convention). Then we have to come up with an idea which real number intervals should be mapped to which integer values. If you want to map [0,0.25[ to 0, [0.25,0.5[ to 1, [0.5,0.75[ to 2 and [0.75,1.0] to 3, the conversion can be done by b = std::floor(f * 4.0) (floor takes only the integer part of a number and ignores the fraction part). This does work for all numbers except f=1. A simple change to b = floor(f >= 1.0 ? 255 : f * 256.0) can fix this problem. This equation ensures that the intervals are equally spaced.

If you assume that our real value is given as a single-precision IEEE 754 floating-point number then there is a finite number of possible float representations within the interval [0,1]. You have to decided which representations of those real numbers belong to which integer representation. Then you can come up with some source code that converts your float number to an integer and check if it fits your mapping. Maybe int ig = int(255.99 * g); is right thing for you or maybe b = floor(f >= 1.0 ? 255 : f * 256.0). It depends on what real number representation you want to map to which integer number representation.

Take a look at the following program. It demonstrates that different conversions do different things:

#include <iostream>

constexpr int realToIntegerPeterShirley(const double value) {
    return int(255.99 * value);

#define F2B(f) ((f) >= 1.0 ? 255 : (int)((f)*256.0))
constexpr int realToIntegerInkredibl(const double value) {
    return F2B(value);

const int realToIntegerMarkByers(const double value) {
    return std::floor(value >= 1.0 ? 255 : value * 256.0);

constexpr int realToIntegerToolmakerSteve(const double value) {
    return std::round(value * 255.0);

constexpr int realToIntegerErichKitzmueller(const double value) {
    return value*255.999;

constexpr int realToInteger(const float value) {
    return realToIntegerInkredibl(value);

int main() {
        double value = 0.906285;
        std::cout << realToIntegerMarkByers(value) << std::endl; // output '232'
        std::cout << realToIntegerPeterShirley(value) << std::endl; // output '231'

        double value = 0.18345;
        std::cout << realToIntegerInkredibl(value) << std::endl; // output '46'
        std::cout << realToIntegerToolmakerSteve(value) << std::endl; // output '47'

        double value = 0.761719;
        std::cout << realToIntegerVertexwahn(value) << std::endl; // output '195'
        std::cout << realToIntegerErichKitzmueller(value) << std::endl; // output '194'

You can use this small testbed to make experiments:

int main() {
    std::mt19937_64 rng;
    // initialize the random number generator with time-dependent seed
    uint64_t timeSeed = std::chrono::high_resolution_clock::now().time_since_epoch().count();
    std::seed_seq ss{uint32_t(timeSeed & 0xffffffff), uint32_t(timeSeed>>32)};
    // initialize a uniform distribution between 0 and 1
    std::uniform_real_distribution<double> unif(0, 1);
    // ready to generate random numbers
    const int nSimulations = 1000000000;
    for (int i = 0; i < nSimulations; i++)
        double currentRandomNumber = unif(rng);

        int firstProposal = realToIntegerMarkByers(currentRandomNumber);
        int secondProposal = realToIntegerErichKitzmueller(currentRandomNumber);

        if(firstProposal != secondProposal) {
            std::cout << "Different conversion with real " << currentRandomNumber << std::endl;
            return -1;

At the end I would suggest not to convert from float to integer. Store your image as high dynamic range data and choose a tool (e.g. that converts your data to low dynamic range. Tone mapping is an own research area and there some tools that have a nice user interface an offer you all kinds of tone map operators.

Image Types, Since indexed images reference color tables composed of up to 256 colors, the data values of Pixel values in an image file can be stored in many different data types. Byte: An 8-bit unsigned integer ranging from 0 to 255. they are usually converted to floating-point or double-precision data types prior to performing� The CSS property to change the color of the text to RGB 0, 0, 255 is called "color". The color property can be set on classes, ids or directly on the HTML element. This example shows how text in the color rgb(0, 0, 255) looks like..text, #text, p{color:rgb(0, 0, 255)}

It is a conversion from "sRGB 0-255" to "linear RGB 0-1". The formula given by Fabian is a simplified approximation which should give good enough results in most cases (it gives larger relative errors in darker regions but that's where the eyes are less sensitive).

Conversion from 0-255 to 0-1 will behave still in the same way, so if you're getting color differences where they should not be, it will IMHO be in the conversion TO 0-255, not FROM it. Again - to "convert" from the range 0 - 255 to 0 - 1, you must divide by 255, that's the only way to get 0 -> 0 and 255 -> 1.

Does the range 0.0-1.0 map to the "standard" 0-255 of 8-bit per channel color, or is the range 0.0-255.0 used for that instead? What are the intended ways of converting from floating p Hi, In HDR images, floating point color is sometimes used.

  • The title of your question says float to byte, but the body of the question says byte to float. Which one do you require?
  • Good point, fixed that. As you can see from the rest of the question it's as it says in title - convert float to byte.
  • BTW, 0.9999 is extremely close to 1.0, and should definitely be converted to 255. Any solution that fails to do so would be wrong.
  • NOTE: Having thoroughly analyzed the math, I've made an in-depth case that round(f * 255.0 is the optimal solution - despite all the answers that are based on * 256 or *255.999. (Though in practice, its usually not significant - the accepted answer's formula is fine. Its also fine to substitute 255.999 for 256 in that answer. My analysis shows that neither of those is optimal - any change from the optimal formula increases the error for some values - but the error increase is minor.)
  • The only thing that concerns me is that now 255 suddenly has a subtly higher range than all the others ;).
  • The "problem" comes from having a closed interval instead of a half-closed interval. There's no way to fix this without having one interval slightly larger than the others. Console yourself by knowing that the distribution of floats in the interval [0,1] is not uniform (they are more densely packed near zero) so there's no guarantee that the other intervals are the same size either.
  • 255 should cover values from 0.99609375 included to 1.0 excluded. This answer suggests to include 1.0 to the interval. Indeed this is very subtle. For me this is the best possible answer.
  • To further clarify: if you start with a closed interval and split it into (say) two intervals, the only thing you can do is to make one half-closed interval and one closed interval. There's no way around the fact that one interval is "slightly larger" than the other. It's best not to worry about it. :)
  • Note: I updated my comment to explicitly (rather than implicitly) call the floor function, in case it is unclear.
  • Agree, it's simpler in code but has less precision than the accepted solution.
  • inkredibl: You will be hard-pressed to find examples where the difference really matters...