Question

如何在以下代码中正确返回set函数中的引用？我确实理解变量set在块结束时超出了范围，但它引用的数组是拥有的并且是活动的。终身'b的尝试是徒劳的。

#[derive(PartialEq, Eq, PartialOrd, Ord, Debug)]
struct Appearance<'a> {
    identity:   &'a u64, 
    role:       &'a str
}

#[derive(Clone, Copy, Debug)]
enum Dereference<'a> {
    Dereference1ary {set: [&'a Appearance<'a>;1]},
    Dereference2ary {set: [&'a Appearance<'a>;2]},
}
impl<'a, 'b> Dereference<'a> {
    fn set(&self) -> &'b [&'a Appearance<'a>] {
        match *self {
            Dereference::Dereference1ary{set} => &set,
            Dereference::Dereference2ary{set} => &set
        }
    }
}

fn main() {
    let r = "hair_color";
    let i1 = 42;
    let i2 = 43;
    let a1 = Appearance{identity: &i1, role: r};
    let a2 = Appearance{identity: &i2, role: r};
    let d1 = Dereference::Dereference1ary{set: [&a1]};
    let d2 = Dereference::Dereference2ary{set: [&a1, &a2]};
    let list: Vec<Dereference> = vec!(d1, d2);
    println!("{:?}", list);
    println!("{:?}", d1.set());
}

这里设有一个游乐场：https://play.rust-lang.org/?gist=fd528ebd054a9ce96004c8608166f9ac&version=stable&backtrace=0

Answer 1

问题是Derefencece1ary{set}复制set并引用该副本。您想要的是使用set

直接引用Dereference1ary{ ref set }字段

impl<'a> Dereference<'a> {
    fn set(&self) -> &[&'a Appearance<'a>] {
        match *self {
            Dereference::Dereference1ary{ref set} => set,
            Dereference::Dereference2ary{ref set} => set,
        }
    }
}

请注意，您的生命周期'b是多余的，但如果您明确要求，则可以写

fn set<'b>(&'b self) -> &'b [&'a Appearance<'a>]

因为你想借用self对象的一部分，所以生命周期需要匹配。

如何返回对拥有和活着的东西的引用？

1 个答案: