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# [derive(Clone)] Is Broken

(rgbcu.be)
128 points RGBCube | 1 comments | 04 Jul 25 20:31 UTC | HN request time: 0s | source
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bloppe ◴[08 Jul 25 08:16 UTC] No.44498108[source]▶
I don't see how "the hard way" is a breaking change. Anybody got an example of something that works now but wouldn't work after relaxing that constraint?
replies(1): >>44498259 #
yuriks ◴[08 Jul 25 08:46 UTC] No.44498259[source]▶
It relaxes the contract required for an existing type with derive(Clone) to implement Clone, which might allow types in existing code to be cloned where they couldn't before. This might matter if precluding those clones is important for the code, e.g. if there are safety invariants being maintained by Type<T> only being clonable if T is clone.
replies(1): >>44502522 #
1. bloppe ◴[08 Jul 25 18:10 UTC] No.44502522[source]▶
Ok let's say there's existing code that requires that a type is not Clone. Then that type definitely would not have #[derive(Clone)] applied to it. So it would not be affected by the change. So it would not be broken.

It's only a breaking change if code that previously worked stops working without changing the code.