当选项可用时,如何展开该选项,否则如何返回默认参考值?[副本]

这是我试图执行的代码:

fn my_fn(arg1: &Option<Box<i32>>) -> i32 {
    if arg1.is_none() {
        return 0;
    }
    let integer = arg1.unwrap();
    *integer
}

fn main() {
    let integer = 42;
    my_fn(&Some(Box::new(integer)));
}

( on the Rust playground )

我在Rust的早期版本中遇到以下错误:

error[E0507]: cannot move out of borrowed content
 --> src/main.rs:5:19
  |
5 |     let integer = arg1.unwrap();
  |                   ^^^^ cannot move out of borrowed content

在更现代的版本中:

error[E0507]: cannot move out of `*arg1` which is behind a shared reference
 --> src/main.rs:5:19
  |
5 |     let integer = arg1.unwrap();
  |                   ^^^^
  |                   |
  |                   move occurs because `*arg1` has type `std::option::Option<std::boxed::Box<i32>>`, which does not implement the `Copy` trait
  |                   help: consider borrowing the `Option`'s content: `arg1.as_ref()`

我看到已经有很多关于借阅检查器问题的文档,但是在阅读之后,我仍然无法找出问题所在。