async fn 调用 split 产生错误 "borrowed value does not live long enough"

Question

我创建了一个人为的示例来理解我在使用 tokio 的异步函数中看到的错误：

pub async fn unique_words_async() -> io::Result<u32> {
    let file = File::open("sample.txt").await?;
    let reader = BufReader::new(file);
    let mut lines = reader.lines();

    let mut h = HashMap::new();
    h.insert("word", true);
    while let Some(line) = lines.next_line().await? {
        let mut split = line.trim().split(' ');
        while let Some(word) = split.next() {
            h.insert(word, true);
        }
    }
    Ok(h.len() as u32)
}

这是确切的错误：

16 |         let mut split = line.trim().split(' ');
   |                         ^^^^^^^^^^^ borrowed value does not live long enough
...
20 |     }
   |     - `line` dropped here while still borrowed
21 |     Ok(h.len() as u32)
   |        ------- borrow later used here

看运行住在playground

完成相同任务的非异步代码工作正常（参见 playground）：

pub fn unique_words() -> u32 {
    let input = "this is one line\nhere is another line";
    let mut lines = input.lines();
    let mut h = std::collections::HashMap::new();
    while let Some(line) = lines.next() {
        let mut split = line.split(' ');
        while let Some(word) = split.next() {
            h.insert(word, true);
        }
    }
    h.len() as u32
}

Answer 1

您的 non-async 翻译不正确。如果将 File 与 BufReader 一起使用，则会出现相同的错误：

pub fn unique_words() -> std::io::Result<u32> {
    use std::io::BufRead;

    let file = std::fs::File::open("sample.txt")?;
    let reader = std::io::BufReader::new(file);
    let mut lines = reader.lines();

    let mut h = std::collections::HashMap::new();
    while let Some(line) = lines.next().transpose()? {
        let mut split = line.split(' ');
        while let Some(word) = split.next() {
            h.insert(word, true);
        }
    }
    Ok(h.len() as u32)
}

error[E0597]: `line` does not live long enough
  --> src/main.rs:10:25
   |
10 |         let mut split = line.split(' ');
   |                         ^^^^^^^^^^^^^^^ borrowed value does not live long enough
...
14 |     }
   |     - `line` dropped here while still borrowed
15 |     Ok(h.len() as u32)
   |        ------- borrow later used here

Playground.

这是因为 Lines::next() returns Option<io::Result<String>>, and the String is dropped at the end of each loop turn. However, str::trim() and str::split() 使用 &str 并生成 &str 具有相同的生命周期（因为它们只对字符串进行切片，而不是更改它）。因此，您将 h 推入一个 &str 并具有一个循环周期的生命周期，即悬空引用。

它与 str::lines() is that std::str::Lines::next() returns Option<&str> 一起工作的原因是引用与原始 &str 相关联，即 'static.

解决这个问题的最简单方法是将 &str 转换为拥有的 String:

h.insert(word.to_owned(), true);

Playground.

async fn 调用 split 产生错误 "borrowed value does not live long enough"

async fn calling split generates error "borrowed value does not live long enough"

asynchronous

rust

rust-tokio

tokio