Rust 捕获FnMut闭包的时间不够长

问题

简而言之:我正在try 实现一个struct，其中包含对闭包的回调，该闭包以静默方式捕获状态.其 idea 是，用户提供回调(闭包)，并且可以在以后发生特定事件时得到通知.

我有一个可用的概念验证版本(code sample #1).

This PoC works (code sample #1个):

fn fun<F>(f: F) -> F
    where
        F: FnMut() -> (),
{
    f
}

fn main() {
    let mut abc = "abc".to_string();
    let mut f = fun(|| {
        abc.push_str(".");
        println!("closure: {:?}", abc);
    });
    f();
    f();
    f();
    println!("end: {:?}", abc);
}

输出:

closure: "abc."
closure: "abc.."
closure: "abc..."
end: "abc..."

This fails (code sample #2个):

与前面的 idea 相同(略有不同)，但试图将闭包包含在Foo中.

struct Foo<'a, T> {
    pub cb: Box<dyn FnMut(&T) + 'a>,
}

impl<'a, T> Foo<'a, T> {
    fn new(f: impl FnMut(&T) + 'a) -> Self {
        Self { cb: Box::new(f) }
    }
    fn on_change(&mut self, f: impl FnMut(&T) + 'a)
    {
        self.cb = Box::new(f);
    }
}

impl<'a, T> Default for Foo<'a, T>
{
    fn default() -> Self {
        Self::new(|_| {})
    }
}

fn main() {
    let mut abc = "abc".to_string();
    let mut f = Foo::default();
    f.on_change(|a| {
        abc.push_str("."); // PROBLEM HERE; uncomment and it works!
        println!("- param: {:?}", a);
    });
    let param = "a".to_string();
    (f.cb)(&param);
    (f.cb)(&param);
    (f.cb)(&param);
    println!("end: {:?}", abc);
}

预期输出:

- param: "a"
- param: "a"
- param: "a"
end: "abc..."

实际输出(编译器错误):

error[E0502]: cannot borrow `abc` as immutable because it is also borrowed as mutable
  --> src/main.rs:33:27
   |
25 |     f.on_change(|a| {
   |                 --- mutable borrow occurs here
26 |         abc.push_str("."); // PROBLEM HERE; uncomment and it works!
   |         --- first borrow occurs due to use of `abc` in closure
...
33 |     println!("end: {:?}", abc);
   |                           ^^^ immutable borrow occurs here
34 | }
   | - mutable borrow might be used here, when `f` is dropped and runs the destructor for type `Foo<'_, String>`

编译器错误是非常明显的，它肯定与生命周期有关.我认为我的问题是我需要告诉编译器闭包和它的参数生命周期-但问题是，如何？

我应该如何修改code sample #2以获得像code sample #1中那样的回调注册工作？

代表:

• code sample #1个
• code sample #2个