rust - 如何不借钱就能确保 self 超越返回值(value)

Question

假设有这样的事情:

trait B {}
struct BB {}
impl B for BB {}

struct A {}
impl A {
    // With this, A can be dropped before B
    // fn get_b(&mut self) -> Box<dyn B> {...}

    fn get_b<'a>(&'a mut self) -> Box<dyn B + 'a> {...}
    fn mut_a(&mut self) {}
}

fn main() {
    let mut a = A {};
    let b = a.get_b();

    // These lines don't compile
    a.mut_a(); // A is borrowed. Is there any way to make this compile?

    // Does not compile and should not. b must be dropped before a
    drop(a);
    drop(b);
}

有什么方法可以确保 a超过 b(出于 unsafe代码的原因)而又不让 a借用？

编辑: a和 b都必须是可变的(如果声明了 mut)并且应该保持可移动状态。它唯一需要确保的是B在A之前被丢弃。

Answer 1

如果给A两个字段，例如lifetime和data，则可以让get_b接受对lifetime的共享引用，让mut_a接受对data的可变引用:

trait B {}
struct BB {}
impl B for BB {}

struct A {
    pub lifetime: ALifetime,
    pub data: AData,
    #[allow(dead_code)]
    private: (), // prevent destructuring
}
struct ALifetime {}
struct AData {}
impl A {
    fn new() -> Self {
        Self { lifetime: ALifetime {}, data: AData {}, private: () }
    }
}
impl ALifetime {
    fn get_b<'a>(&'a self) -> Box<dyn B + 'a> { Box::new(BB{}) }
}
impl AData {
    fn mut_a(&mut self) {}
}

fn main() {
    let mut a = A::new();
    let b = a.lifetime.get_b();
    a.data.mut_a(); // Only `a.data` is borrowed here.
    // drop(b);
    drop(a); // Not allowed. b must be dropped before a
}

游乐场: https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=ccd1a25cacd21d4364f2f3add1499015