rust - 两个使用对父作用域的引用的闭包

Question

我试图创建一个存储 2 个闭包的结构，每个闭包都具有对创建它们的父作用域变量的读取引用。

在我关闭了父作用域中的引用之后，我如何确保它们与闭包存在相同的时间？

例如:

struct S {
    x: bool,
}

type Closure = Box<dyn Fn()>;

struct Env {
    fn_a: Closure,
    fn_b: Closure,
}

fn get_env() -> Env {
    let s = S { x: true };

    let fn_a = || {
        &s.x;
    };
    let fn_b = || {
        &s.x;
    };

    Env {
        fn_a: Box::new(fn_a),
        fn_b: Box::new(fn_b),
    }
}

fn main() {
    let env = get_env();
}

playground

结果:

error[E0597]: `s` does not live long enough
  --> src/main.rs:16:10
   |
15 |     let fn_a = || {
   |                -- value captured here
16 |         &s.x;
   |          ^ borrowed value does not live long enough
...
23 |         fn_a: Box::new(fn_a),
   |               -------------- cast requires that `s` is borrowed for `'static`
...
26 | }
   | - `s` dropped here while still borrowed```

Answer 1

这取决于你的场景。返回的闭包会存在多久？

如果它的生命周期与 Env 一样长值(value)，搬家怎么样s进入Env值和访问 s来自 Env值(value)？

如果您不知道(或者不想费心去弄清楚)闭包的生命周期，一个方便(虽然效率稍低)的方法是使用 Rc , 然后 move结果中 Rc 的克隆:

Playground

use std::rc::Rc;

pub struct S {
    x: bool,
}

pub type Closure = Box<dyn Fn()>;

#[allow(unused)]
pub struct Env {
    fn_a: Closure,
    fn_b: Closure,
}

fn get_env() -> Env {
    let s = Rc::new(S { x: true });

    let cloned = Rc::clone(&s);
    let fn_a = move || {
        s.x;
    };
    let fn_b = move || {
        cloned.x;
    };

    Env {
        fn_a: Box::new(fn_a),
        fn_b: Box::new(fn_b),
    }
}

fn main() {
    let _env = get_env();
}