将struct转换为具有lifetime got的特征“由于要求冲突，无法推断lifetime参数`'a`的适当生存期”

Question

trait BT {
    fn get_a(&self) -> &A;
}

#[derive(Debug)]
struct A {
    v: i32,
}

impl A {
    fn nb(&self) -> Box<BT> {
        Box::new(B { a: self })
    }
}

#[derive(Debug)]
struct B<'a> {
    a: &'a A,
}

impl<'a> BT for B<'a> {
    fn get_a(&self) -> &A {
        return self.a;
    }
}

fn main() {
    println!("{:?}", A { v: 32 }.nb().get_a());
}

A有一个方法来生成一个引用为A的B实例，并且B可能有许多方法访问B.a (A在B中的引用)。如果让A.nb()返回B而不是BT，代码会工作得很好。

我是Rust的新手。这个问题整天都在困扰着我。我该怎么做才能让这段代码工作呢？谢谢!

整个错误报告：

error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'a` due to conflicting requirements
  --> src\socket\msg\message.rs:53:26
   |
53 |                 Box::new(B{a: self})
   |                          ^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 52:13...
  --> src\socket\msg\message.rs:52:13
   |
52 | /             fn nb(&self) -> Box<BT> {
53 | |                 Box::new(B{a: self})
54 | |             }
   | |_____________^
note: ...so that reference does not outlive borrowed content
  --> src\socket\msg\message.rs:53:31
   |
53 |                 Box::new(B{a: self})
   |                               ^^^^
   = note: but, the lifetime must be valid for the static lifetime...
   = note: ...so that the expression is assignable:
           expected std::boxed::Box<socket::msg::message::test::test::BT + 'static>
              found std::boxed::Box<socket::msg::message::test::test::BT>

Answer 1

特征对象的默认生存期是'static。您需要为nb()函数返回的特征对象添加一个显式的生命周期绑定：

impl A {
    fn nb<'s>(&'s self) -> Box<BT+'s> {
        Box::new(B{a: self})
    }
}

Inference of Trait Object Lifetimes