有没有办法允许移动具有借用元素但不丢弃它的容器？

Question

我有这个容器：

use std::ptr::NonNull;

struct Container {
    data: NonNull<u8>,
}

impl Container {
    fn new() -> Container {
        todo!()
    }
    fn borrow_some_heap_data<'a>(&'a self) -> &'a u8 {
        todo!()
    }
    fn borrow_some_heap_data_mut<'a>(&'a mut self) -> &'a mut u8 {
        todo!()
    }
}

impl Drop for Container {
    fn drop(&mut self) {
        todo!()
    }
}

fn main() {
    let container = Container::new();
    let data = container.borrow_some_heap_data(); // or mut

    {
        let container = container; // move

        // This is safe because the move is a memcpy and the heap data doesn't change.
        println!("{}", *data);
    }

    // This is not safe because the container has been dropped
    // println!("{}", *data);
}

error[E0505]: cannot move out of `container` because it is borrowed
  --> src/main.rs:30:25
   |
27 |     let data = container.borrow_some_heap_data(); // or mut
   |                --------- borrow of `container` occurs here
...
30 |         let container = container; // move
   |                         ^^^^^^^^^ move out of `container` occurs here
...
33 |         println!("{}", *data);
   |                        ----- borrow later used here

即使有引用，移动容器也是安全的。然而，丢弃它并不安全。有什么方法可以在 Rust 中表达这一点，允许移动但不允许 drops?

data 在结构被删除之前永远不会被释放。

Answer 1

不，在安全的 Rust 中无法做到这一点；编译器不够智能/没有语法来区分容器的生命周期和元素的生命周期。

来自：

There is a special case where the lifetime tracking is overzealous: when you have something placed on the heap. This occurs when you use a Box<T>, for example. In this case, the structure that is moved contains a pointer into the heap. The pointed-at value will remain stable, but the address of the pointer itself will move. In practice, this doesn't matter, as you always follow the pointer.

有没有办法允许移动具有借用元素但不丢弃它的容器？

Is there any way to allow moving a container that has a borrowed element but not dropping it?

collections

rust

borrow-checker