alt.hn

6/1/2026 at 5:37:20 PM

Should you normalize RGB values by 255 or 256?

 https://30fps.net/pages/255-vs-256-division/

by pplanu

6/1/2026 at 11:01:31 PM

This problem of what exactly a color value means is mostly inconsequential when you have 8 bits per component, the difference in the denominator being either 255 or 256 makes the errors tiny, you must have really good color perception and get really close to the screen to see any difference at all, and your monitor/phone screen is probably not calibrated anyway, so who cares.

It becomes a pain in the ass when you're generating a VGA signal with a microcontroller with 8 color output pins (3 red, 3 green, 2 blue). The meaning of a color value is very real in this setup: it corresponds to a voltage level you must send to the VGA monitor, 0V-0.7V.

So the blue channel will map (0->0V, 1->0.23V, 2->0.47V, 3->0.7V), and the red/green will map (0->0V, 1->0.1V, ..., 7->0.7V). Notice how none of the blue voltages match any of the red/green ones (other than the extremes)? That means you don't get to see any pure grays -- the closest ones will have bit of blue or yellow tint, depending on the direction of the difference.

Not only that, any gradients at all (other than the ones not mixing blue with the other channels) will be noticeable off: for example, the closest colors in the line between pure red to pure white will all be slightly orange or purple.

Code for VGA output in 8-bit color with double-buffered 320x240 framebuffer for the Raspberry Pi Pico 2 here, if anyone cares: https://github.com/moefh/pico-vga-8bit-demo

by moefh

6/2/2026 at 2:02:32 AM

>our monitor/phone screen is probably not calibrated anyway,

I'm not convinced this holds nowadays, the higher end/flagship phones seem to have commonly have calibrated screens nowadays. Usually they have a high saturation profile loaded as default, but the 'realistic' profile usually turns out to be pretty well calibrated. Similarly with laptops and monitors. I've tried using a colorimeter on my recent devices only to find that the display was already calibrated pretty well.

by hgoel

6/2/2026 at 12:50:44 AM

You forgot about gamma correction. Before converting a value in the range of 0-255 into a voltage, PCs typically raise that value to the power of 2.2. This makes the difference between small values and large values far more apparent:

2^2.2 = 4.595, 255^2.2 = 196,964.699

by chongli

6/2/2026 at 12:31:41 AM

There is a fallacy here in assuming there's 256 steps from 0 to 255. That's not true, there's 256 values that can be represented in 8 bits, and 255 steps (spaces between those values) from 0 (black) to (255) pure white. Thus, the division by 255 isn't problematic. Of course, 128 isn't half grey, it isn't in 0-255 and quantized 8-bit values are almost always in sRGB, not linear perceptive space.

This is the same kind confusion that happens with sampling positions in modern APIs, where the location is specified in coordinates and not in pixel centers.

by BearOso

6/1/2026 at 7:04:20 PM

I'll argue for the +0.5 solution. First, I don't like half-sized intervals at the edges, and second, a 255-based representation is typically a SDR (not HDR) image.

RGB values represent luminances against some adapted state, and a "zero" in a daylit scene is not "zero luminance" - it's just about 0.001x as bright as the brightest point - it's millions of photons, way more than zero. In a sense our eyes experience contrast on a sliding scale, and there is no absolute zero in the system. For example, broadcast systems historically used 16-235 as their luminance range for SDR. I think any argument that says "we must have zero" is going to have a bias, but I don't think zero is needed for most things.

by herf

6/1/2026 at 8:54:31 PM

As someone with a lot of experience in this area doing image processing and rendering for VFX (including writing image readers and writers for my own software and commercial VFX software), I think you might be forgetting that colourspace conversion (to sRGB 'linear' rec709 for old-school SDR, but other more wider gamuts for newer formats) would happen after this, so the 'squish' of the dynamic range would happen after loading.

Also, a lot of workflows for image processing and compositing do assume that 0 means zero, whether correctly or not (often incorrectly). So there are often assumptions that for 8-bit, 0u maps to 0.0f and 255 maps to 1.0f for things like masking or alpha: as soon as you have 0 values which become just over 0.0, you then have artifacts because some code somewhere is using a hard threshold of 0.0 to mask some other operation, and vice-versa for 1.0 with alpha, where suddenly because the 255 values are no longer 1.0f, you have very slightly see-through objects (often only visible in certain situations or when pixel-peeping) after pre-multiplication.

(Same thing can happen when 254 becomes 1.0f after +0.5 with masking).

by pixelesque

6/1/2026 at 9:32:08 PM

good point - alpha is a notable exception, it is not luminance

by herf

6/1/2026 at 10:43:57 PM

Although the post focuses on RGB, the same quantization issue exists for any type of signal being mapped between discrete and continuous representations.

The issue isn't in having a representation for 0 photons, but about maximizing information stored in a byte. Ideally you shouldn't be underutilizing the byte value 0, nor add bias to data that should have been assigned to the 0th bucket, regardless of what it represents (you could have a color space that goes from bright to super bright, and still want to ensure that every byte represents equal chunk of your brightness range).

by pornel

6/1/2026 at 11:53:49 PM

Yep, the exact same problem arises in digital audio, mapping between integer sample formats and the floating point representation that is generally used internally.

by PaulDavisThe1st

6/1/2026 at 7:14:33 PM

Both solutions add 0.5, the difference is where in the process it happens.

by yxhuvud

6/1/2026 at 7:50:22 PM

I agree. Additionally, both 0.0 and 1.0 don't really exist for dithered signals, so a byte should map to [0.5, 255.5] before division by 256. This also solves the signed integer asymmetry, as a signed byte maps to [-127.5, 127.5] before division by 128. I wonder if audio DSP folks have done this already.

by amavect

6/1/2026 at 10:22:59 PM

Thinking about this more, dithering requires negative values to cancel out when adding. Works for audio, but color doesn't have negative numbers.

by amavect

6/1/2026 at 11:18:56 PM

It is still frequency, where it would have negative values. but I doubt any color handling algorithms deal with it as a frequency. Rightfully so, the physical wetware for decoding images is very different than that for decoding audio. Well... not that different if you think of audio as a single pixel monochrome image.

Now I am imagining a weird alternate history where we treat audio like we treat color. OK take three bytes which encode how loud the sound is, one for lows, one for mids and one for highs where lows mids and high frequencies are picked to match human ear response.

by somat

6/1/2026 at 11:27:21 PM

They are not half sized at the edges, unless negative black bothers you.

by kazinator

6/1/2026 at 8:45:45 PM

Interesting idea, but somehow I feel the world is shaking. For the processing program, what used to black(0.0) and white(1.0) has became very dark gray and very bright gray.

by infinet

6/1/2026 at 10:10:44 PM

> broadcast systems historically used 16-235

For 8-bit, 16 maps to 7.5IRE which is the well understood legal black. Mapping 235 means they mapped peak to 110IRE. This is based on a 0-120IRE scale. This gets weird as the broadcast limit for video was 100IRE allowing for the chroma to reach 110IRE. So if you're trying to limit your white values to 235, that'll be higher than is broadcast safe. Of course, nobody cares about NTSC broadcast limits any more. However, to this day, I still see out of spec tapes marked as "broadcast master" that have been ingested for streaming use. It drives me crazy to this day, and it's only getting worse as people don't even have scopes to adjust the VTR's TBC properly.

by dylan604

6/2/2026 at 12:14:43 AM

> For 8-bit, 16 maps to 7.5IRE which is the well understood legal black. Mapping 235 means they mapped peak to 110IRE.

Generally no -- in an 8-bit NTSC-M Rec. 601 system, 16 maps to E'Y = 0 at 7.5 IRE, and 235 maps to E'Y = 1 at 100 IRE. See https://www.poynton.ca/pdf/Poynton-1996-TechIntrDigiVide.pdf

The "16" digital black level is independent of the "7.5 IRE" analog setup. E.g. in Japan with an 8-bit "NTSC-J" Rec. 601 system, my understanding is that 16 still maps to E'Y = 0 which is now at 0 IRE, and 235 is still E'Y = 1 at 100 IRE.

by keithwinstein

6/1/2026 at 11:36:40 PM

Ugh. Sudden flashbacks to having to switch analog output between Japanese NTSC (no pedestal) and US NTSC (with pedestal) without getting weird noise in the black regions.

But IIRC the MPEG-2 standard had luma==235 -> 100IRE for all of the analog formats (pal/ntsc-j/ntsc/secam) so I'm not sure why you say that would violate the broadcast limits?

by variaga

6/1/2026 at 8:12:52 PM

> In a sense our eyes experience contrast on a sliding scale

There's a whole visual center to check the amount of incoming light and adjust your pupils for you. It's intentionally reactive.

> and there is no absolute zero in the system.

There maybe is. I think we call that "blind."

> broadcast systems historically used 16-235 as their luminance range for SDR

Mostly because it was a fully analog system and these all translate down to signal voltage. Jokingly NTSC used to be referred to as "Never Twice the Same Color" due to being a compromise bolted onto the side of an already compromised system.

by themafia

6/1/2026 at 9:59:41 PM

>> and there is no absolute zero in the system.

> There maybe is. I think we call that "blind."

If you go looking into that, you'll see that the reality is far far more complex [0]

"The number of people with no light perception is unknown, but it is estimated to be less than 10 percent of totally blind individuals."

[0] https://chicagolighthouse.org/sandys-view/what-blind-people-...

by a_conservative

6/1/2026 at 8:39:46 PM

That was a fun article to read of something I haven't had to think about in a while. It brought to mind moments in game development of having pixel art needing to be drawn on an integer value despite the game logic using floating point math. I tried something similar to the +0.5 in places so that it wouldn't look as bad (especially when there's a moving camera, which also needed to be truncated..).

I also enjoyed the 2002 article by Jonathan Blow [1] that's linked at the bottom. The visualization from the first article helped a lot once this started to go more in-depth.

[1] https://web.archive.org/web/20240706043551/https://number-no...

by Nuthen

6/1/2026 at 6:56:28 PM

If you have a ruler and it goes to 12 inches, you should normalize by the length L and not by 13, the number of points on the ruler.

by dudu24

6/1/2026 at 8:54:12 PM

I'm confused by that analogy. Is the “ruler” a 255-inch ruler with 256 points labeled 0–255, or is it a 256-inch ruler with 256 1-inch segments, making L = 256×1?

by Timwi

6/2/2026 at 12:44:27 AM

the correct way is to use a slide rule

by m463

6/1/2026 at 10:12:55 PM

But who says that the numbers are representing the points, rather than representing the intervals between the points?

by layer8

6/1/2026 at 10:38:50 PM

It doesn't even need to represent intervals. A 13 inch ruler with 13 markings at 0.5, 1.5, etc inches is still a valid ruler, albeit an odd construction.

by wky

6/1/2026 at 7:10:19 PM

yes but >> 8 is so much faster

by lacedeconstruct

6/1/2026 at 8:14:47 PM

You don’t divide a float by 256 by shifting it right eight bits; that would yield complete garbage. You subtract 8 from the exponent, then check if you got an underflow.

by xigoi

6/1/2026 at 9:05:57 PM

Same point; divide by power of 2 is a fast subtraction operation in float world, while divide by 255 shits all over the whole float

by dheera

6/1/2026 at 11:56:05 PM

If your input is an arbitrary float, you need to check for denormals (and maybe NaNs). You can do bitmasking trick to avoid conditional jumps but I'm skeptical you can do it faster than SIMD multiply instruction.

by yongjik

6/1/2026 at 8:06:24 PM

It's just multiplication. Floating multiply is extraordinarily fast.

by StilesCrisis

6/1/2026 at 8:13:22 PM

The difference between 20 cycles and 1 clock cycle in a hot loop is very noticeable

by lacedeconstruct

6/1/2026 at 8:46:50 PM

FP Division by constant is optimized by a compiler into a multiply. Graphics processing typically happens on the GPU these days, and on all recent GPUs FPMUL belongs to the class of lowest-latency operations. That is, there are no other instructions that complete faster.

by Tuna-Fish

6/1/2026 at 9:17:56 PM

Only with things like -ffast-math enabled will compilers do the reciprocal. It can make a fair difference in some cases, but it's often better to selectively use it in code locations you know are acceptable by doing it manually in the code.

by pixelesque

6/1/2026 at 9:03:25 PM

That's only valid to do if the reciprocal is representable exactly.

by mgaunard

6/1/2026 at 10:14:49 PM

That's not totally true. It's sufficient to be exactly representable, but you only need the reciprocal rounding error to be small enough to guarantee the multiplication rounding step fixes it across the entire range of numerators. For IEEE754 f16 values, there are 28 such extra values, the positive and negative sides of 1705/x where x is a power of 2 at least as great as 2048.

by hansvm

6/1/2026 at 8:35:09 PM

Useful, then, that you can start several vectorized floating-point muls each cycle. (E.g., most modern x86 are 3/0.5 cycles for vmulps. No 20 cycles in sight.)

by Sesse__

6/1/2026 at 7:41:20 PM

Only in micro-benchmarks.

For real usage, today's CPUs are limited by memory bandwidth.

by dist-epoch

6/1/2026 at 7:51:01 PM

What are you talking about in a hot loop in my software renderer this is like 10x faster

    // color4_t result = {
    //     .r = (src.r * src.a + dst.r * inv_alpha) * INV_255,
    //     .g = (src.g * src.a + dst.g * inv_alpha) * INV_255,
    //     .b = (src.b * src.a + dst.b * inv_alpha) * INV_255,
    //     .a = src.a + (dst.a * inv_alpha) * INV_255
    // };

    // 1/256 but much faster
    color4_t result = {
        .r = (src.r * src.a + dst.r * inv_alpha) >> 8,
        .g = (src.g * src.a + dst.g * inv_alpha) >> 8,
        .b = (src.b * src.a + dst.b * inv_alpha) >> 8,
        .a = src.a + ((dst.a * inv_alpha) >> 8)
    };

by lacedeconstruct

6/1/2026 at 8:53:44 PM

If the latter is 10x faster, the issue is some kind of weird compilation failure for the above version. For one, it only cuts a third of the multiplies.

by Tuna-Fish

6/1/2026 at 7:51:38 PM

Because you are working in the cache.

Also, you should use SIMD.

by dist-epoch

6/1/2026 at 7:59:03 PM

> Also, you should use SIMD. ironically no clang is better at auto vectorizing

by lacedeconstruct

6/1/2026 at 7:43:28 PM

[dead]

by szundi

6/1/2026 at 7:11:16 PM

I’m dumb. Doesn’t 0 start at the beginning?

by groundzeros2015

6/1/2026 at 10:16:03 PM

It's right up there with the confusion if 2000 was the new year of the 21st century or the last year of the 19th century.

by dylan604

6/1/2026 at 11:01:41 PM

For the record, the mathematically correct answer to this question is that the year 2000 was the last year of the 19th century.

The reason is that year 0 never existed. The year 1 BCE was followed by the year 1 CE.

Culturally, anthropologically, and psychologically it might be a different matter. But 2000 years had not passed before the end of that year.

by simonask

6/1/2026 at 11:29:48 PM

The debate is if 2000 is the first year of the 21st century or the last year of the 20th century. (btw I agree with the latter)

by tzot

6/2/2026 at 12:55:37 AM

wow, yeah, that's quite the miss on my part.

by dylan604

6/1/2026 at 11:49:22 PM

The entire issue arises from the use of truncation, right? It guarantees that only an exact 1.0 could land in the 255 bin so the net effect is a reduction of 256 bins to 255 bins. (Using random numbers as shown also guarantees no 1.0.)

Why not scale to fill the available bins, though? i.e. trunc(result * 255.999)?

by dpark

6/2/2026 at 12:29:17 AM

> Why not scale to fill the available bins, though? i.e. trunc(result * 255.999)?

That’s half of the mid-riser staircase quantizer discussed in the article. (The other half is coming up with the reverse.)

(I would implement it as min(floor(x * 256), 255).)

by klodolph

6/2/2026 at 1:11:06 AM

Kind of. According to the article, the mid-riser vs mid-tread distinction correlates to where the 0.5 is applied in the transform. I’m proposing that there is no 0.5 applied at all. Instead of counteracting the compression of the truncation operation by adding a fixed offset, it multiplies by a scale factor.

Possibly my proposal doesn’t hold up to repeated transforms and operations. It might skew toward 255 in real operations.

by dpark

6/2/2026 at 1:03:15 AM

There are only two real solutions after factoring in the need to preserve black as zero.

They are "rgb / 255.0" vs. "rgb / 256.0". Both have different tradeoffs. Pick your poison. (If you're using a 8 bit display signal then you better match whatever value the OS picked for the mapping back to the display, so your RGB values pass through unchanged)

by 2001zhaozhao

6/1/2026 at 11:26:51 PM

No, the "alternative" approach looks strange in the 7 bit example.

1.0 lies on the right side of the bin 7. But 0.0 lies on the left of bin 0.

The standard approach assumes that we have centered samples: that zero is dead black, plus (and minus!) some uncertainty and so is bin 7.

If the sampling of the intensity is distortion-free (no clipping took place due to overexposure) then bin 7 represents a range of possible values centered around 1.0.

It is not a half-sized interval.

> This means that when converting floating-point values in the [0,1] range back to integers, the extreme bins have effectively half the width of other bins.

Under any interpretation whatsoever of the image samples, there is latitude for interpreting the maximum value 255 as being distortion: clipping from an arbitrarily higher value. Shifting things by 0.5 doesn't fix this issue of not knowing whether 255 means that an intensity close to 1.0 is being represented (no distortion), or an outlier intensity of 37.49 (severely clamped). That could go the other way too.

In other words, there is a possible bias in the extreme bin. The signal could be limited such that the bin's full sampling range is not in effect, or the signal could be overwhelming, so that values far outside of the range are clipped and included.

The only way around this is to make the highest value a canary which represents "clipped value". That is to say, 255 means "clipped datum", so that only 254 and below is sampling of unclipped signal. Machine-generated image (e.g. 3D rendering) then avoid the 255 value, and camera sensors are calibrated so that it doesn't occur when technical images are being shot.

by kazinator

6/1/2026 at 7:54:33 PM

You should multiply by 255.0, optionally add a dither (triangular is okay), and then let the FPU round using its default IEEE 754 round-to-nearest-ties-to-nearest-even mode. None of this crazy 0.5 stuff. :-)

by Sesse__

6/1/2026 at 9:24:40 PM

The author is confusing bins with bin edges. In their first plot, the standard approach looks strange because 0-7 should be the bin edges, not the center points as shown in the plot.

You can see this confusion again in the histogram example. There are only 255 bins, not 256. If you fix that mistake and remove the 0.5 offset, then the histogram is distributed correctly at both ends.

by jessetemp

6/1/2026 at 10:33:20 PM

No, the author understands the problem way deeper than you do.

You haven't grasped the fact that the choice isn't obvious, and has subtle trade-offs.

If you don't believe the author, check the other posts he references.

by pornel

6/1/2026 at 9:58:51 PM

2*8 = 256. You can represent 256 distinct values, bins, with an 8 bit number. If you stick a 0 in that first one, it takes a bin. If you fill the rest with by-one increasing integers, then the max value will be 255, thus the 2*bits - 1, which is the max value you can store.

by nomel

6/1/2026 at 9:48:32 PM

How do you fit 256 distinct values into 255 bins?

by bjourne

6/1/2026 at 10:17:30 PM

By counting the edges

by jessetemp

6/2/2026 at 1:13:39 AM

I see what you’re saying - index 0 holds values from 0-1, index 2 from 1-2 etc, but then you have index 255 holding values between 255 and 256. So you’re sort of arguing that the 0-255 8-bit quantization is actually representing ‘real’ values of 0-256?…

Edit: somehow missed alterom’s reply - they explain it much better than my question above does.

by jrmg

6/2/2026 at 12:32:31 AM

Sorry, you seem to be confused.

>There are only 255 bins, not 256

There are 256 bins because there are 256 values.

The questions are:

1. What are the boundaries of these bins?

2. Which sample represents a particular bin?

With 1-bit color, we have sample values {0, 1}. What bins do they represent?

Here's one choice:

     [0, 1), [1, 2)
Two equally sized bins, spanning the interval [0, 2] of length 2, each defined by its sample at lower bound.

Alternatively, we could consider these bins:

    [-0.5, 0.5), [0.5, 1.5)
These are also equally sized bins, spanning the interval [-0.5, 1.5] of length 2, defined by samples at the center.

We could also define bins like this:

    [0, 0.5), [0.5, 1]
Two equally sized bins spanning the interval [0, 1] of length 1, where we sample the first bin at the lower bound, and the last bin at the upper bound.

This, in a nutshell is what the author is trying to explain.

Let's look at this again, with 2 bits.

With 2-bit color, we have sample values {0, 1, 2, 3}.

Which bins do they come from?

The three options above yield:

    [0, 1), [1, 2), [2, 3), [3, 4)

    [-.5, 0.5), [0.5, 1.5), [1.5, 2.5), [2.5, 3.5)

    [0, 0.5), [0.5, 1.5), [1.5, 2.5), [2.5, 3]
The first two span an interval of length 4, the third spans an interval of length 3.

In the third case, the tail bins are short (have size ½), and the rest have size 1.

The last bin must be a closed interval in the third case, so that it includes the value we picked to represent it.

None of these choices is inherently invalid or better than the others; and none stems from "confusing bins with edges".

The third option does have the distinction that the first and last bins are smaller than the rest. But it's not necessarily a drawback. Especially when we're talking about color, hardware interpretation, and human perception.

When you remap these bins into the [0, 1] interval, you're "dividing by 4" in the first two cases, and by 3 in the third case.

The maps are:

     x → x/4
     x → (x + ½)/4
     x → x/3
The inverse maps (that yield a sample in {0, 1, 2, 3} given a floating point value in interval [0, 1]) are:

    x → trunc(4x)
    x → round(4x - ½) = trunc(4x)
    x → trunc(3x + ½)
In the first two options, the domain is [0, 1). It might be necessary to apply clipping because the exact value 1.0 falls outside the range of the forward transform.

The 2nd option is the most symmetric, of course, but the 3rd one is the most straightforward (and cheapest) to implement, so that's the default.

The choice amounts to making the highest and lowest bins slightly smaller to make the rest sightly larger.

That's to say, if you generate uniform noise between 0 and 1, you'll get the following samples from your function with equal probability:

    0 or 3
    1
    2
As the author points out, this hardly matters when you are talking about having 256 bins.

That, and with color specifically, the "good" histograms aren't uniform anyway (and any photographer wants to avoid getting much at either extreme).

TL;DR: The author is not confusing anything — but their diagram and explanation are, indeed, a bit confusing.

by alterom

6/1/2026 at 7:19:33 PM

Both of these assume a linear transfer function, which is rarely the case.

by Retr0id

6/1/2026 at 8:34:14 PM

Basically never for 8-bit color channels.

by leni536

6/1/2026 at 7:21:23 PM

Advice for anyone on mobile: read in landscape mode if you want to be able to see the division by 256 version code example at the start.

The HTML/CSS is bad that lets it completely overflow the right edge of the page instead of wrapping.

I re-read this post three times in total confusion before I figured out the most important piece was off-screen entirely.

by crazygringo

6/1/2026 at 7:05:05 PM

Should always be 0-255 as that fits an unsigned byte.

by theyeenzbeanz

6/1/2026 at 7:26:23 PM

That's not what the article is about.

by crazygringo

6/1/2026 at 7:23:39 PM

> assume that in both cases the output values are clamped before the final typecast

by Retr0id

6/1/2026 at 7:56:57 PM

Interesting article. I tend to use

- i = min(floor(f * 256), 255) (from float to uint8)

- f = i / 255 (from uint8 to float)

Basically a mix of the 2 approaches mentioned in the article.

For all integers between [0,255], if I do uint8 -> float -> uint8 conversion, I will get the same result.

--

edit: I wondered what's the maximum jitter amount that I can introduce to the float and get the same uint8 value. And also these 0->0.0 and 255->1.0 should map properly.

With my approach at the top, the jitter margin that I can introduce is 1/65280.

But with the article's approach

- i = floor(f * 255 + 0.5)

- f = i / 255

maximum jitter margin is 1/510 (which is better).

by atilimcetin

6/1/2026 at 8:48:58 PM

It's worth pointing out that the article explicitly calls out your first mixed technique:

> Finally, one should never mix the encode and decode steps of the two quantizers. That’s just broken code. It’s an easy mistake to make, though.

by AgentME

6/1/2026 at 8:16:34 PM

This is what I do for the former:

    floor( nextafter( 256, 255 ) * value )

by vitorsr

6/1/2026 at 8:19:10 PM

Oh very nice idea to get rid of the min operator.

by atilimcetin

6/1/2026 at 10:46:11 PM

Are we talking 0 or 1 based values? HONKHONK*

by RobRivera

6/1/2026 at 10:50:43 PM

Case against 255: it looks wrong in the graph :(

Case against 256: no 0 or 1 values :(

Considering how important having a 0 and 1 value is for arithmetic in general, I think 255 is better.

by AlienRobot

6/1/2026 at 7:37:06 PM

A similar issue exists in the audio world, for example 16-bit integer audio is between [-32768, 32767] (non-symmetric), but floating point audio is [-1.0, 1.0].

by dist-epoch

6/1/2026 at 11:26:26 PM

There is an analogous situation in graphics with signed normalized formats. The solution there is that the R16_SNORM format maps -1 to +1 as [-32767, 32767] with -32768 being a special value (not normally emitted, and mostly but not always interpreted as -32767). Some audio storage formats seem to use this mapping too.

by ack_complete

6/1/2026 at 7:44:14 PM

note that floating point audio very often exceeds [-1.0, 1.0] within the pipeline, just to be tamed at the very end of the mix to fit within those bounds. this is pretty much why every modern DAW uses floating point these days.

by adzm

6/1/2026 at 11:42:55 PM

You don't need to make this judgement; it's fixed by the colorspace you're working in.

First, figure out what colorspace the processing needs to happen in. Usually this is linear RGB.

Then, figure out what OETF and EOTF your input/output format use. This will be something like PQ or HLG. This will exactly specify the meaning of each integer value.

This fixes the choice of representation and conversion.

by wyager

6/1/2026 at 11:51:27 PM

[flagged]

by davidladdsource

6/1/2026 at 9:53:29 PM

[dead]

by corysama

6/1/2026 at 7:55:53 PM

Both. 255 for each color and the last 1 as the alpha for each channel.

Why not??? Fight me

by ctdinjeu8

6/1/2026 at 6:55:09 PM

255 gives 0-255, which gives you a zero value. 256 is 1-256, you lose the option of setting 0.

by DigitallyFidget

6/1/2026 at 7:27:47 PM

That's not what the article is about.

by crazygringo

6/1/2026 at 8:18:51 PM

"Let’s say you’re writing an image processing program. The program takes in an image, converts it to floating point, does some processing and finally saves the modified pixels to disk as 8-bit colors. "

excuse to argue about the best way aside, if this is the goal you should not be rolling your own image file reading. you should use openimageio. idk what approach it takes in its internal conversion to float, but that library is more likely to have the right answer than you trying to roll it yourself given its the library used internally by tons of professional image manipulation software...

by dgently7

6/1/2026 at 8:41:01 PM

If you're a beginner, or just want something which works quickly, sure.

However OIIO is far from perfect in all situations (having had to debug and fix issues with its mip-map generation filtering code in the past), so don't always assume that just because there's a mature open source library out there doing something that it's always perfect.

by pixelesque

6/1/2026 at 9:17:51 PM

sure of course nothing is perfect and oiio has a lot of surface area / is still oss. thats good advice.

ive just seen a lot of "ai researchers" who are getting into professional image processing and are both beginners and want things quickly and so could do much worse than just starting from what they get out of oiio. especially for a lot of the non-obvious stuff (more of that in color handling than just the io stuff though)

by dgently7