rust - 我如何在这里返回＆str？

Question

当我尝试运行此命令时:

use std::thread::LocalKey;

struct Person {
    name: String,
}

impl Person {
    fn new(name: String) -> Self {
        Person {
            name: name,
        }
    }
    
    fn name(&self) -> &str {
        self.name.as_str()
    }
}

thread_local! {
    static NAMED_THING: Person = Person::new("John".to_string());
}

fn retrieve_name(p: &'static LocalKey<Person>) -> &str {
    p.with(|n| n.name())
}

fn main() {
    println!("The name is {}", retrieve_name(&NAMED_THING));
}

它告诉我

error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
  --> src/main.rs:24:18
   |
24 |     p.with(|n| n.name())
   |                  ^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #2 defined on the body at 24:12...
  --> src/main.rs:24:12
   |
24 |     p.with(|n| n.name())
   |            ^^^^^^^^^^^^
note: ...so that reference does not outlive borrowed content
  --> src/main.rs:24:16
   |
24 |     p.with(|n| n.name())
   |                ^
   = note: but, the lifetime must be valid for the static lifetime...
note: ...so that reference does not outlive borrowed content

我可以改为:

fn retrieve_name(p: &'static LocalKey<Person>) -> &str {
    p.with(|n| n.name().to_owned().as_str())
}

但是每次都克隆该字符串似乎很浪费。更不用说，它现在告诉我:

error[E0515]: cannot return value referencing temporary value
  --> src/main.rs:24:16
   |
24 |     p.with(|n| n.name().to_owned().as_str())
   |                -------------------^^^^^^^^^
   |                |
   |                returns a value referencing data owned by the current function
   |                temporary value created here

Playground

Answer 1

你不能做这个。
当线程结束时，Person将被销毁，然后对其的任何引用(例如，与此同时已发送到另一个线程的'static引用)将变得悬而未决。这就是LocalKey具有with方法的原因，而不是像Deref变量那样实现lazy_static的原因:因为将'static引用引用到线程局部值是不合理的。
您需要组织代码，以便所有借用都限于传递给with的闭包范围。例如:

fn print_name(p: &'static LocalKey<Person>) {
    p.with(|n| println!("The name is {}", n.name()));
}

fn main() {
    print_name(&NAMED_THING);
}