When encountering `<&T as Clone>::clone(x)` because `T: Clone`, suggest `#[derive(Clone)]`
CC #40699.
```
warning: call to `.clone()` on a reference in this situation does nothing
--> $DIR/noop-method-call.rs:23:71
|
LL | let non_clone_type_ref_clone: &PlainType<u32> = non_clone_type_ref.clone();
| ^^^^^^^^
|
= note: the type `PlainType<u32>` does not implement `Clone`, so calling `clone` on `&PlainType<u32>` copies the reference, which does not do anything and can be removed
help: remove this redundant call
|
LL - let non_clone_type_ref_clone: &PlainType<u32> = non_clone_type_ref.clone();
LL + let non_clone_type_ref_clone: &PlainType<u32> = non_clone_type_ref;
|
help: if you meant to clone `PlainType<u32>`, implement `Clone` for it
|
LL + #[derive(Clone)]
LL | struct PlainType<T>(T);
|
```
86727df4ed