我可以通过保留引用来避免克隆字符串吗?

我有一个数据结构,在这个结构中,我在读取缓冲区周围提供了一个包装器,以自动处理读出器中的重复语句。

这是通过存储剩余重复次数和要重复的行的内部状态来实现的。

use std::fs::File;
use std::path::Path;
use std::io::BufReader;
use std::io::prelude::*;
use std::io::Error;
use std::num::NonZeroU32;
use std::mem;

pub struct Reader {
    handle: BufReader<File>,
    repeat_state: Option<(NonZeroU32, String)>,
}

impl Reader {
    pub fn new<P: AsRef<Path>>(path: P) -> Result<Reader, Error> {
        let file = File::open(path)?;
        let handle = BufReader::new(file);

        Ok(Reader {
            handle,
            repeat_state: None,
        })
    }

    /// get next line, respecting repeat instructions
    pub fn next_line(&mut self) -> Option<String> {
        if self.repeat_state.is_some() {
            let (repeats_left, last_line) = mem::replace(&mut self.repeat_state, None).unwrap();

            self.repeat_state = NonZeroU32::new(repeats_left.get() - 1)
                .map(|repeats_left| (repeats_left, last_line.clone()));

            Some(last_line)
        } else {
            let mut line = String::new();
            if self.handle.read_line(&mut line).is_err() || line.is_empty() {
                return None
            }

            if line.starts_with("repeat ") {
                let repeats: Option<u32> = line.chars().skip(7)
                    .take_while(|c| c.is_numeric())
                    .collect::<String>().parse().ok();

                self.repeat_state = repeats
                    .and_then(|repeats| NonZeroU32::new(repeats - 1))
                    .map(|repeats_left| (repeats_left, line.clone()))
            }

            Some(line)
        }
    }
}

#[test]
fn test_next_line() {
    let source = "
line one
repeat 2    line two and line three
line four
repeat 11   lines 5-15
line 16
line 17
last line (18)
    ".trim();
    let mut input = File::create("file.txt").unwrap();
    write!(input, "{}", source);


    let mut read = Reader::new("file.txt").unwrap();
    assert_eq!(
        read.next_line(),
        Some("line one\n".to_string())
    );
    assert_eq!(
        read.next_line(),
        Some("repeat 2    line two and line three\n".to_string())
    );
    assert_eq!(
        read.next_line(),
        Some("repeat 2    line two and line three\n".to_string())
    );
    assert_eq!(
        read.next_line(),
        Some("line four\n".to_string())
    );

    for _ in 5..=15 {
        assert_eq!(
            read.next_line(),
            Some("repeat 11   lines 5-15\n".to_string())
        );
    }

    assert_eq!(
        read.next_line(),
        Some("line 16\n".to_string())
    );
    assert_eq!(
        read.next_line(),
        Some("line 17\n".to_string())
    );
    assert_eq!(
        read.next_line(),
        Some("last line (18)".to_string())
    );
}

Playground

&str . 我试过几种方法,但都没能成功:

• 储存 String ,返回 &str公司 :“活得不够长”终身错误
• 储存 &str公司 &str公司 :相同的生存期错误
• Cow<&str>
• Box<&str>

这些克隆是我的程序目前的瓶颈,根据CodeXL基于时间的采样探查器,在构建了带有调试信息的发布模式之后。现在,我的程序虽然很快,但我想知道是否有办法避免它们。