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### What it does
Checks for usage of `&Option<&T>`.
### Why is this bad?
Since `&` is Copy, it's useless to have a
reference on `Option<&T>`.
### Known problems
It may be irrelevant to use this lint on
public API code as it will make a breaking change to apply it.
### Example
```
let x: &Option<&u32> = &Some(&0u32);
```
Use instead:
```
let x: Option<&u32> = Some(&0u32);
```
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