rust - 您能否在一个位置更新一个数字并在引用的任何地方进行更改？

Question

我想编写一个程序，该程序将具有许多(也许是几十到数百个)对相同数字的引用，例如“x”，并且每秒将被更新多次。因此，我只想在一个位置上更改编号，并使所有引用自动更新，而不必重新分配它们(再次，因为可能有数百个引用，导致数百次重新分配。)。如果我可以只在一个地方更新数字，那么我可以立即运行数百个product()更新。
下面是我对此的尝试，其中有两个结构引用同一f64内存。我只想更新一次x，然后更新我的结构(thing1和Thing2)的所有实例，以正确计算product。

#[derive(Debug)]
struct Thing<'a> {
    high: &'a f64,

    low: &'a f64,
}

impl Thing<'_> {
    pub fn product(&self) -> f64 {
        *self.high * *self.low
    }
}

fn main() {
    let mut x: f64 = 42.01;
    let y: f64 = 7.95;
    let w: f64 = 99.9;
    let mut thing1: Thing = Thing { high: &w, low: &x };
    let mut thing2: Thing = Thing { high: &y, low: &x };
    println!("Before {:?} {}", thing1, thing1.product());
    println!("Before {:?} {}", thing2, thing2.product());

    // I want to swap or copy in a new value for the reference x, because
    // it appears multiple times in different `Thing`s.
    x = 500.7_f64;
    println!("After {:?} {}", thing1, thing1.product());
    println!("After {:?} {}", thing2, thing2.product());
}

上面的代码无法编译并出现以下错误:

error[E0506]: cannot assign to `x` because it is borrowed
  --> src/main.rs:25:5
   |
18 |     let mut thing1: Thing = Thing { high: &w, low: &x };
   |                                                    -- borrow of `x` occurs here
...
25 |     x = 500.7_f64;
   |     ^^^^^^^^^^^^^ assignment to borrowed `x` occurs here
26 |     println!("After {:?} {}", thing1, thing1.product());
   |                               ------ borrow later used here

Answer 1

您可以使用 Cell 及其表亲 RefCell ，用于修改共享值的用例:

use std::cell::Cell;

#[derive(Debug)]
struct Thing<'a> {
    high: &'a Cell<f64>,
    low: &'a Cell<f64>,
}

impl Thing<'_> {
    pub fn product(&self) -> f64 {
        self.high.get() * self.low.get()
    }
}

fn main() {
    let x = Cell::new(42.01);
    let y = Cell::new(7.95);
    let w = Cell::new(99.9);
    let thing1: Thing = Thing { high: &w, low: &x };
    let thing2: Thing = Thing { high: &y, low: &x };
    println!("Before {:?} {}", thing1, thing1.product());
    println!("Before {:?} {}", thing2, thing2.product());

    x.set(500.7_f64);
    println!("After {:?} {}", thing1, thing1.product());
    println!("After {:?} {}", thing2, thing2.product());
}

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