如何在返回新接收的值和缓存值之间的所有组合的迭代器时修复生命周期问题？

Posted 2021-04-21

我试图返回一个新接收的值和缓存值之间的所有组合的迭代器，但我有一个终身问题。

use std::collections::HashMap;

pub struct Status {
    // <i,Vec<u>>
    old: HashMap<usize, Vec<usize>>,
}

impl<'a> Status {
    pub fn new() -> Self {
        Status {
            old: HashMap::new(),
        }
    }

    pub fn gen(&mut self, u: usize, i: usize) -> UserPairIter<'a> {
        let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
        UserPairIter::new(entry, u)
    }
}

struct UserPairIter<'a> {
    data: &'a mut Vec<usize>,
    u: usize,
    index: usize,
}

impl<'a> UserPairIter<'a> {
    pub fn new(data: &'a mut Vec<usize>, u: usize) -> Self {
        UserPairIter { data, u, index: 0 }
    }
}

impl<'a> Iterator for UserPairIter<'a> {
    type Item = (usize, usize);

    fn next(&mut self) -> Option<Self::Item> {
        if self.index >= self.data.len() {
            self.data.push(self.u);
            return None;
        }
        let result = (self.u, self.data[self.index]);
        self.index += 1;
        Some(result)
    }
}

fn main() {}

当我编译时，我收到以下错误消息：

error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
  --> src/main.rs:16:50
   |
16 |         let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
   |                                                  ^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 15:5...
  --> src/main.rs:15:5
   |
15 | /     pub fn gen(&mut self, u: usize, i: usize) -> UserPairIter<'a> {
16 | |         let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
17 | |         UserPairIter::new(entry, u)
18 | |     }
   | |_____^
note: ...so that reference does not outlive borrowed content
  --> src/main.rs:16:41
   |
16 |         let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
   |                                         ^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 8:1...
  --> src/main.rs:8:1
   |
8  | / impl<'a> Status {
9  | |     pub fn new() -> Self {
10 | |         Status {
11 | |             old: HashMap::new(),
...  |
18 | |     }
19 | | }
   | |_^
note: ...so that reference does not outlive borrowed content
  --> src/main.rs:16:41
   |
16 |         let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
   |                                         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

答案

首先，不要把生命周期放在Status中。如果您需要函数的通用性，请将此参数放在函数中：

impl Status {
    pub fn new() -> Self {
        Status {
            old: HashMap::new(),
        }
    }

    pub fn gen<'a>(&mut self, u: usize, i: usize) -> UserPairIter<'a> {
        let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
        UserPairIter::new(entry, u)
    }
}

然后编译器说它无法推断self.old的生命周期。只需给它&'a mut self提示，以便编译器理解UserPairIter的生命周期与Status的生命周期相同：

impl Status {
    pub fn new() -> Self {
        Status {
            old: HashMap::new(),
        }
    }

    pub fn gen<'a>(&'a mut self, u: usize, i: usize) -> UserPairIter<'a> {
        let entry: &'a mut Vec<usize> = self.old.entry(i).or_insert(Vec::new());
        UserPairIter::new(entry, u)
    }
}

那没关系！

你不需要说entry的类型，编译器可以用函数签名推断它：

pub fn gen<'a>(&'a mut self, u: usize, i: usize) -> UserPairIter<'a> {
    let entry = self.old.entry(i).or_insert(Vec::new());
    UserPairIter::new(entry, u)
}