为什么我的变量不够长寿？

Question

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() {}

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 valid for the anonymous lifetime #1 defined on the function body at 4:1...
  --> src/main.rs:4:1
   |
4  | / fn file_to_vec(filename: &str) -> Result<Vec<&str>, io::Error> {
5  | |     let mut file = try!(File::open(filename));
6  | |     let mut string = String::new();
7  | |     try!(file.read_to_string(&mut string));
...  |
12 | |     Ok(data)
13 | | }
   | |_^

为什么我一直收到这个错误？我该如何解决？我想这与split方法有关.

我可以返回字符串然后将其拆分为Vecmain函数中的a,但我真的想要返回一个向量.

Answer 1

问题是string在函数中创建,并在函数返回时被销毁.要返回的向量包含切片string,但这些切片在函数外部无效.

如果您对性能不是非常担心,可以Vec<String>从功能中返回.您只需返回类型Result<Vec<String>, io::Error>并更改该行

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

至

let data: Vec<String> = string.split('\n').map(String::from).collect();