Question

我想保留一个Holded实例，但我不能，因为它在一个数组中。如何从数组中“提取”此实例并将其保存在Object实例中？（在我的原始代码中，我没有数组但是迭代器）。这是等效的代码：

struct Holded {
    value: u8,
}

struct Holder;

impl Holder {
    pub fn get(&self) -> [Holded; 2] {
        [Holded { value: 0 }, Holded { value: 1 }]
    }
}

struct Object<'a> {
    holded: &'a Holded,
}

fn main() {
    let holder = Holder;

    let obj = work(&holder).unwrap();
    println!("{}", obj.holded.value);
}

fn work(holder: &Holder) -> Option<Object> {
    let mut obj: Object;
    let array = holder.get();

    for h in array.into_iter() {
        if h.value == 1u8 {
            obj = Object { holded: h };
            return Some(obj);
        }
    }

    None
}

错误消息：

error: `array` does not live long enough
  --> src/main.rs:28:14
   |
28 |     for h in array.into_iter() {
   |              ^^^^^ does not live long enough
...
36 | }
   | - borrowed value only lives until here
   |
note: borrowed value must be valid for the anonymous lifetime #1 defined on the block at 24:43...
  --> src/main.rs:24:44
   |
24 | fn work(holder: &Holder) -> Option<Object> {
   |                                            ^

Answer 1

在您发布的MCVE中，结构Object包含对Holded的引用：

struct Object<'a> {
    holded: &'a Holded,
}

在函数work()中，您返回Object（可选）：

fn work(holder: &Holder) -> Option<Object> {

您从一个按值返回的函数中获取Holded：

impl Holder {
    pub fn get( &self ) -> [Holded; 2] {
        [Holded { value: 0 }, Holded { value: 1 }]
    }
}

现在这将永远不会奏效。如果您返回对Holded的引用，则您引用的Holded必须存储在某处。这意味着要么作为输入，要么作为函数work()的输出。

我重写了您的示例，将Holded包含在Holder中。这是解决这个问题的一种方法。但我不确定这是否适用于您原来的问题。

struct Holded {
    value: u8,
}

struct Holder{
    value: [Holded; 2],
}

impl Holder {
    pub fn new() -> Holder {
        Holder {value: [Holded { value: 0 }, Holded { value: 1 }] } 
    }

    pub fn get( &self ) -> &[Holded; 2] {
        &self.value
    }
}

struct Object<'a> {
    holded: &'a Holded,
}

fn main() {
    let holder = Holder::new();

    let obj = work(&holder).unwrap();
    println!("{}", obj.holded.value);

    let obj = work2(&holder).unwrap();
    println!("{}", obj.holded.value);
}

fn work(holder: &Holder) -> Option<Object> {
    let obj: Object;
    let array = holder.get();

    for h in array.into_iter() {
        if h.value == 1u8 {
            obj = Object { holded: h };
            return Some(obj);
        }
    }

    None
}

fn work2(holder: &Holder) -> Option<Object> {
    holder.get()
        .iter()
        .filter(|h| h.value == 1u8)
        .map(|h| Object { holded: h })
        .next()
}

您注意到我还添加了一种不同的方式来实现work()函数（work2()）。

`array`的寿命不够长

1 个答案: