Borrow Checker 3 @BoxyUwU

Warning: this quiz is still "work-in-progress", some questions might not have good explanations (or any at all), formatting/structure/titles/etc are not final, and so on. You might want to return here on a later date.

fn foo(a: &mut u32) -> u32 {
    let mut b = || &mut *a;

    *b() = 12;
    *b() = 73;
    *a
}

fn main() {
    foo(&mut 292);
}
Solution 
error: captured variable cannot escape `FnMut` closure body
 --> examples/borrowck_3.rs:2:20
  |
1 | fn foo(a: &mut u32) -> u32 {
  |        - variable defined here
2 |     let mut b = || &mut *a;
  |                  - ^^^^^^-
  |                  | |     |
  |                  | |     variable captured here
  |                  | returns a reference to a captured variable which escapes the closure body
  |                  inferred to be a `FnMut` closure
  |
  = note: `FnMut` closures only have access to their captured variables while they are executing...
  = note: ...therefore, they cannot allow references to captured variables to escape

error: could not compile `code` (example "borrowck_3") due to 1 previous error

The closure assigned to b can only return a reference with one specific lifetime (because it has no lifetimes in arguments). So both calls to b return a unique (mutable) reference with the same lifetime, which is an error, because all unique references must be disjoint.

If you inline calls to b it would work though:

fn foo(a: &mut u32) -> u32 {
    let mut b = || &mut *a;

    *(&mut *a) = 12;
    *(&mut *a) = 73;
    *a
}

This is fine because reborrows return different lifetimes, allowing them to be disjoint.