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Is an observable prop, but not observable? #3579

Answered by hitsthings
hitsthings asked this question in Q&A
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Yeah I forgot the annotation for notes in the first version, but had edited it in - didn't affect the results. isComputedProp` is what I actually wanted, thanks!

I think my misunderstanding is rooted in the fact that observableArray::filter does not return an observable array - just a regular array. I will have to keep that in mind in the future. Thanks for being part of my learning process here. :)

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