How to convert a PreAlpha into [u8; 4]?

[suguruwataru]

Firstly, thank you for the great work!

The question is in the title. The use case is to convert data read from image directly into a premultiplied memory buffer for further processing. I feel it's a pretty common use case but I can't find an obvious way of doing it. There are more convoluted ways of doing this, like directly computing on the public member variables or taking advantage of that PreAlpha Derefs into the Rgb it wraps (Rgb only converts to [u8; 3] and alpha needs to be added back), but I want to know whether there is a method that is more concise and idiomatic.

[Ogeon]

Hi, thanks for the kind words, I'm glad it's appreciated. 😊

You are right, there's currently no direct support built in. The snag, apart from a lack of an into_format method, is that the struct itself only accepts non-integers (i.e. anything that can be used in the multiplication process) as things are implemented. There's no particular reason for this, other than that it seemed to make sense at the time, so feel free to create an issue for it. There may be a solution that's not a breaking change. If not, I think it makes sense to relax the requirement in the next major release.

[suguruwataru]

I see, thank you for replying. It's a great crate, you certainly deserve all the kind words ;)

[Ogeon]

I added #450 as a reminder to improve it.

[suguruwataru]

That's great! Thank you so much

[clouds56]

Besides PreAlpha only accepts Scalar: Real, I found Alpha<C, T> and PreAlpha<C> differs so that Alpha<C, T> might have different type between C and T, like Alpha<Rgb<S, u8>, f32> is valid while PreAlpha<C> would have Scalar determined by C.
Which one is better design in your opinion?

[Ogeon]

Alpha has its additional type parameter so you can process the color part separately from the transparency part. For example, something like the CSS style Alpha<Srgb<u8>, f32>. It may be a good idea to make PreAlpha support that too, for the same reason.