相关疑难解决方法(0)

为什么我的变量不够长寿?

我有一段简单的代码,可以按行将文件读入矢量

use std::io::{self, Read};
use std::fs::File;

fn file_to_vec(filename: &str) -> Result<Vec<&str>, io::Error> {
    let mut file = try!(File::open(filename));
    let mut string = String::new();
    try!(file.read_to_string(&mut string));
    string.replace("\r", "");

    let data: Vec<&str> = string.split('\n').collect();

    Ok(data)
}

fn main() {}
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我收到以下错误:

error[E0597]: `string` does not live long enough
  --> src/main.rs:10:27
   |
10 |     let data: Vec<&str> = string.split('\n').collect();
   |                           ^^^^^^ does not live long enough
...
13 | }
   | - borrowed value only lives until here
   |
note: borrowed value must be …
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lifetime rust

0
推荐指数
1
解决办法
991
查看次数

无法将字符串拆分为具有显式生存期的字符串切片,因为字符串的活动时间不够长

我正在写一个应该从实现BufRead特性的东西中读取的库; 网络数据流,标准输入等.第一个函数应该从该读取器读取数据单元并返回一个填充的结构,该结构主要填充有&'a str从线路中的帧解析的值.

这是一个最小版本:

mod mymod {
    use std::io::prelude::*;
    use std::io;

    pub fn parse_frame<'a, T>(mut reader: T)
    where
        T: BufRead,
    {
        for line in reader.by_ref().lines() {
            let line = line.expect("reading header line");
            if line.len() == 0 {
                // got empty line; done with header
                break;
            }
            // split line
            let splitted = line.splitn(2, ':');
            let line_parts: Vec<&'a str> = splitted.collect();

            println!("{} has value {}", line_parts[0], line_parts[1]);
        }
        // more reads down here, therefore the reader.by_ref() above …
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rust

0
推荐指数
1
解决办法
553
查看次数

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rust ×2

lifetime ×1