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Re: [css-color] Remove gray() Notation

From: Christoph Päper <christoph.paeper@crissov.de>
Date: Tue, 19 Aug 2014 10:43:25 +0200
Message-Id: <C9B0ECDF-FC6A-4E8B-BB77-581CB503D92B@crissov.de>
To: www-style list <www-style@w3.org>
Patrick Dark 2014-08-19:
> 
> You seem to be claiming that rgb(50%, 50%, 50%)/rgb(50%) is more intuitive than hsl(0deg, 0%, 50%)/hsl(50%) because you can delete and add two arguments from the end, instead of the beginning, of the function value to arrive at a grayscale value.

Where “parameters” are optional in CSS, these are usually not the first ones and often they take the same value as the one provided (although sometimes they take their default value), e.g. vertical-horizontal position (where a=aa), TRBL-distances (where a=aaaa, ab=abab, abc=abcb) and hex color notation (‘#abc’ = ‘#aabbcc’). 

‘rgb(50%)’ follows this nicely, because all three values are the same, but ‘hsl(50%)’ would be less intuitive indeed. The percentage cannot denote the hue, because that’s conventionally treated as an angle, but without deeper knowledge it’s unclear whether the percentage specifies saturation or lightness or both. It‘s, of course, just specifying lightness for HSL, but if I remember correctly it would be both, saturation and brightness/value, for HSB/HSV (which are not supported in CSS, yet).

That being said …

> Ergo, HSL-based grayscale value shortcuts are too unintuitive to be worthwhile.

… I’m not sure that notation was *too* unintuitive, since it’s not actually counter-intuitive.
Received on Tuesday, 19 August 2014 08:43:55 UTC

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