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clone_from behaviour for ManuallyDrop #86288
Description
With #[derive(Clone)] on ManuallyDrop<T>, x.clone_from(&y) will not drop the T inside x, just like x = y.clone(); does not drop the original x before overwriting it.
However, a manual implementation of clone_from can only make use of the efficiency of a overriden clone_from on T (e.g. memory reuse by a Vec::clone_from) by simply forwarding to it, which means that x.clone_from(&y) will not forget the old value.
This new (forwarding, so (potentially) dropping) behaviour was added in #85176, and (for now) reverted in #85758.
So question for @rust-lang/libs: What behaviour would be correct? There are arguments for both. Whatever the decision, we should document it and add a test for it.
Activity
Amanieu commentedon Jun 15, 2021
I feel that the old behavior is correct (not calling drop).
clone_fromshould behave likex = y.clone()which forManuallyDropmeans not dropping the old value.I don't think there is much loss of efficiency in practice: use of
clone_fromis already pretty rare and I don't think there are any cases of it being used withManuallyDrop.m-ou-se commentedon Jul 5, 2021
Sounds good. Closing this.
m-ou-se commentedon Jul 5, 2021
No wait, we still need to do this part. :)
Rollup merge of rust-lang#86880 - m-ou-se:test-manuallydrop-clone-fro…