callback - 如何将带有捕获的可变变量的回调像正常的可变借用一样对待？

Question

Foo 可以使用方法 .modify() 修改:

struct Foo;
impl Foo {
    fn modify(&mut self) {}
}

Bar 存储回调:

struct Bar<'a> {
    callback: Box<FnMut() + 'a>,
}
impl<'a> Bar<'a> {
    fn new<F: FnMut() + 'a>(f: F) -> Bar<'a> {
        Bar {
            callback: Box::new(f),
        }
    }
}

init() 获取一片 Bar 并执行它们的回调:

fn init(bars: &mut [Bar]) {
    for b in bars {
        (*b.callback)();
    }
}

现在最有趣的是:

在循环中更改 Foo 效果很好；在循环的每次迭代中，可变地借用 foo 并调用 .modify():

fn main() {
    let mut foo = Foo;

    for _ in 0..10 {
        foo.modify();
    }
}

在回调中更改 Foo 不起作用:

fn main() {
    let mut foo = Foo;

    let mut bar1 = Bar::new(|| foo.modify());
    let mut bar2 = Bar::new(|| foo.modify());

    init(&mut [bar1, bar2]);
}

Try it on the playground ,它有一个错误:

error[E0499]: cannot borrow `foo` as mutable more than once at a time
  --> src/main.rs:27:29
   |
26 |     let mut bar1 = Bar::new(|| foo.modify());
   |                             -- --- previous borrow occurs due to use of `foo` in closure
   |                             |
   |                             first mutable borrow occurs here
27 |     let mut bar2 = Bar::new(|| foo.modify());
   |                             ^^ --- borrow occurs due to use of `foo` in closure
   |                             |
   |                             second mutable borrow occurs here
...
30 | }
   | - first borrow ends here

如何对第2项进行类似的保证？

Answer 1

您可以使用 RefCell :

let foo = RefCell::new(Foo);

{
    let bar1 = Bar::new(|| foo.borrow_mut().modify());
    let bar2 = Bar::new(|| foo.borrow_mut().modify());
    init(&mut [bar1, bar2]);
}

let mut foo = foo.into_inner(); // extract foo to use in external API

小心 borrow_mut()，如果值当前被借用，它会发生 panic 。

---

如果可以更改 Bar 和 init()，则可以将值 foo 传递给 init() 与方法 modify() 分开:

struct Bar<'a> {
    callback: Box<FnMut(&mut Foo) + 'a>,
}
impl<'a> Bar<'a> {
    fn new<F: FnMut(&mut Foo) + 'a>(f: F) -> Bar<'a> {
        Bar {
            callback: Box::new(f),
        }
    }
}

fn init(bars: &mut [Bar], arg: &mut Foo) {
    for b in bars {
        (*b.callback)(arg);
    }
}

let mut bar1 = Bar::new(|x| x.modify());
let mut bar2 = Bar::new(Foo::modify); // you can pass it without closure
init(&mut [bar1, bar2], &mut foo);