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Add a note about scoping differences between *x and *x.deref() #2073

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24 changes: 24 additions & 0 deletions src/expressions/operator-expr.md
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Original file line number Diff line number Diff line change
Expand Up @@ -189,6 +189,30 @@ let y = &mut 9;
assert_eq!(*y, 11);
```

> [!NOTE]
> The [temporary scopes][temporary scope] of `x` in `*x` and `*std::ops::Deref::deref(&x)` are not always the same in all immutable place expression contexts.
> Likewise, the temporary scopes of `x` in `*x` and `*std::ops::DerefMut::deref_mut(&mut x)` are not always the same in all mutable place expression contexts.
>
> If `*x` has an [extended][temporary lifetime extension] temporary scope, then `x` does as well ([destructors.scope.lifetime-extension.sub-expressions]).
> However, the same does not apply to `*std::ops::Deref::deref(&x)` or `*std::ops::DerefMut::deref_mut(&mut x)`.
>
> For example:
>
> ```rust
> // The temporary holding the result of `String::new()` is extended
> // to live to the end of the block, so `x` may be used in subsequent
> // statements.
> let x = &*String::new();
> # x;
> ```
>
> ```rust,compile_fail,E0716
> // The temporary holding the result of `String::new()` is dropped at
> // the end of the statement, so it's an error to use `y` after.
> let y = &*std::ops::Deref::deref(&String::new()); // ERROR
> # y;
> ```

r[expr.try]
## The try propagation expression

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