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Hello! This answer might be obvious but I'm having trouble figuring it out. Can a single For context, I need to end up with a list of unique entities that had a given component type added or updated. If 2 observers are indeed required, is the best approach to sum their lists, sort, and remove duplicates? |
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Answered by
skypjack
Sep 16, 2023
Replies: 1 comment 4 replies
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Uhm, out of my mind, it should be |
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4 replies
Answer selected by
Net5F
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Uhm, out of my mind, it should be
entt::collector.group<T>().update<T>()
. Have you tried it already?