我正在尝试存储一个可变向量元素的引用，以供以后使用。然而，一旦我改变了向量，我就不能再使用存储的引用了。我理解这是因为借用对元素的引用也需要借用对向量本身的引用。因此，不能修改vector，因为这需要借用一个可变引用，而当另一个对vector的引用已经被借用时，这是不允许的。
这里有一个简单的例子

struct Person {
    name: String,
}

fn main() {
    // Create a mutable vector
    let mut people: Vec<Person> = ["Joe", "Shavawn", "Katie"]
        .iter()
        .map(|&s| Person {
            name: s.to_string(),
        })
        .collect();

    // Borrow a reference to an element
    let person_ref = &people[0];

    // Mutate the vector
    let new_person = Person {
        name: "Tim".to_string(),
    };
    people.push(new_person);

    // Attempt to use the borrowed reference
    assert!(person_ref.name == "Joe");
}

这会产生以下错误

error[E0502]: cannot borrow `people` as mutable because it is also borrowed as immutable
  --> src/main.rs:21:5
   |
15 |     let person_ref = &people[0];
   |                       ------ immutable borrow occurs here
...
21 |     people.push(new_person);
   |     ^^^^^^^^^^^^^^^^^^^^^^^ mutable borrow occurs here
...
24 |     assert!(person_ref.name == "Joe");
   |             --------------- immutable borrow later used here

我还尝试过将向量元素装箱，如建议的here，但这没有帮助。我以为它可以让我删除对vector的引用，同时保持对元素的引用，但显然不是。

struct Person {
    name: String,
}

fn main() {
    // Create a mutable vector
    let mut people: Vec<Box<Person>> = ["Joe", "Shavawn", "Katie"]
        .iter()
        .map(|&s| {
            Box::new(Person {
                name: s.to_string(),
            })
        })
        .collect();

    // Borrow a reference to an element
    let person_ref = people[0].as_ref();

    // Mutate the vector
    let new_person = Box::new(Person {
        name: "Tim".to_string(),
    });
    people.push(new_person);

    // Attempt to use the borrowed reference
    assert!(person_ref.name == "Joe");
}

这仍然会产生相同的错误

error[E0502]: cannot borrow `people` as mutable because it is also borrowed as immutable
  --> src/main.rs:23:5
   |
17 |     let person_ref = people[0].as_ref();
   |                      ------ immutable borrow occurs here
...
23 |     people.push(new_person);
   |     ^^^^^^^^^^^^^^^^^^^^^^^ mutable borrow occurs here
...
26 |     assert!(person_ref.name == "Joe");
   |             --------------- immutable borrow later used here

有没有办法做到这一点，或者我试图做一些不可能的事情？