[css-color-4] Does interpolation with achromatic colors truly have a "longer" arc?

[facelessuser]

This question is in relation to the recently closed issue: #9224.

Based on the conversation there, it appears it was determined that Chrome was handling logic for interpolation with undefined hues correctly. So this question is to specifically clarify the point in the title.

I realize the original question was specifically asking about "transparent" colors, but the resolution there I believe translates to achromatic colors in general. My personal opinion is that the resolution leads to unexpected results.

Consider the following example which interpolates a red color with black.

linear-gradient(to right in oklch longer hue, oklch(0.628 0.26 29.23), oklch(0 0 none / 1));

Results in Chrome

https://codepen.io/facelessuser/pen/MWZPmwy

Because the undefined hue is resolved to inherit the hue of the red color before the hue fix-up, we now get an interpolation around the entire color wheel, as if there was an arc between black and red. The reality is that achromatic colors have no hue; therefore, they have no arc between any hue.

Based on the conversation in #9224, my understanding is that this is indeed "correct" for CSS, even though it seems unintuitive, at least to me.

While Safari may be wrong in its implementation, it seems intuitive. In their case, it seems the undefined hue is not resolved until after the hue fix-up.

So, this is simply to clarify whether this behavior, conceptually, is truly the intention of the CSS spec or not, outside of which browser implemented the spec correctly.

[svgeesus]

So this:

looks like a correct implementation if shorter was specified. The hues are the same and the interpolation curve in Oklch is a straight line from red to black.

If shorter evaluates to zero degrees of hue difference, then clearly longer will evaluate to 360 degrees - the long way round.

[facelessuser]

Okay, so achromatic hues are treated as having an arc then. That answers my question, even if I disagree. Thanks!