相关疑难解决方法(0)

我有一个值,我想在我自己的类型中存储该值以及对该值内部内容的引用:

struct Thing {
    count: u32,
}

struct Combined<'a>(Thing, &'a u32);

fn make_combined<'a>() -> Combined<'a> {
    let thing = Thing { count: 42 };

    Combined(thing, &thing.count)
}

有时候,我有一个值,我想在同一个结构中存储该值和对该值的引用:

struct Combined<'a>(Thing, &'a Thing);

fn make_combined<'a>() -> Combined<'a> {
    let thing = Thing::new();

    Combined(thing, &thing)
}

有时,我甚至没有参考该值,我得到同样的错误:

struct Combined<'a>(Parent, Child<'a>);

fn make_combined<'a>() -> Combined<'a> {
    let parent = Parent::new();
    let child = parent.child();

    Combined(parent, child)
}

在每种情况下,我都会收到一个错误,即其中一个值"活不够长".这个错误是什么意思？

我有一个可观察的集合和一个观察者.我希望观察者成为一个特质实现trait Observer.当某些事件发生时,可观察对象应该能够通知每个观察者.这应该解释我的意图:

struct A {
    observables: Vec<Observable>,
}

impl A {
    fn new() -> A {
        A {
            observables: vec![],
        }
    }
}

trait Observer {
    fn event(&mut self, _: &String);
}

impl Observer for A {
    fn event(&mut self, ev: &String) {
        println!("Got event from observable: {}", ev);
    }
}

struct Observable {
    observers: Vec<dyn Observer>, // How to contain references to observers? (this line is invalid)
}

impl Observable {
    fn new() -> Observable {
        Observable …